00 0672324539 fm 8/28/03 9:35 AM Page i **Robert Lafore**

Data Structures

& Algorithms

in Java

Second Edition

800 East 96th Street, Indianapolis, Indiana 46240

00 0672324539 fm 8/28/03 9:35 AM Page ii **Data Structures and Algorithms in Java, **

**Executive Editor**

**Second Edition**

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00 0672324539 fm 10/10/02 9:13 AM Page iii Contents at a Glance

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

**1**

Overview. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

**2**

Arrays . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

**3**

Simple Sorting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

**4**

Stacks and Queues . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115

**5**

Linked Lists . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179

**6**

Recursion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251

**7**

Advanced Sorting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 315

**8**

Binary Trees. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 365

**9**

Red-Black Trees . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 429

**10**

2-3-4 Trees and External Storage. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 463

**11**

Hash Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 519

**12**

Heaps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 579

**13**

Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 615

**14**

Weighted Graphs. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 669

**15**

When to Use What . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 717

**Appendixes**

**A**

Running the Workshop Applets and Example Programs . . . . . . . . . . . . . . . . . . . . 729

**B**

Further Reading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 735

**C**

Answers to Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 739

Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 749

00 0672324539 fm 10/10/02 9:13 AM Page iv Table of Contents

**Introduction 1**

What’s New in the Second Edition .............................................................1

Additional Topics ...............................................................................1

End-of-Chapter Questions .................................................................2

Experiments ........................................................................................2

Programming Projects ........................................................................2

What This Book Is About ............................................................................2

What’s Different About This Book ..............................................................3

Easy to Understand ............................................................................3

Workshop Applets ..............................................................................4

Java Example Code .............................................................................5

Who This Book Is For ..................................................................................5

What You Need to Know Before You Read This Book ................................5

The Software You Need to Use This Book ...................................................6

How This Book Is Organized .......................................................................6

Enjoy Yourself! .............................................................................................8

**1**

**Overview 9**

What Are Data Structures and Algorithms Good For? ...............................9

Real-World Data Storage ..................................................................10

Programmer’s Tools ..........................................................................11

Real-World Modeling .......................................................................11

Overview of Data Structures ......................................................................11

Overview of Algorithms ............................................................................12

Some Definitions .......................................................................................13

Database ...........................................................................................13

Record ...............................................................................................13

Field ..................................................................................................13

Key ....................................................................................................14

Object-Oriented Programming ..................................................................14

Problems with Procedural Languages ..............................................14

Objects in a Nutshell ........................................................................15

A Runnable Object-Oriented Program .............................................18

Inheritance and Polymorphism .......................................................21

Software Engineering .................................................................................21

00 0672324539 fm 10/10/02 9:13 AM Page v Java for C++ Programmers .........................................................................22

No Pointers .......................................................................................22

Overloaded Operators ......................................................................25

Primitive Variable Types ...................................................................25

Input/Output ....................................................................................26

Java Library Data Structures ......................................................................29

Summary ....................................................................................................30

Questions ...................................................................................................30

**2**

**Arrays 33**

The Array Workshop Applet ......................................................................33

Insertion ...........................................................................................35

Searching ..........................................................................................36

Deletion ............................................................................................36

The Duplicates Issue .........................................................................37

Not Too Swift ....................................................................................39

The Basics of Arrays in Java .......................................................................39

Creating an Array .............................................................................40

Accessing Array Elements .................................................................40

Initialization .....................................................................................41

An Array Example ............................................................................41

Dividing a Program into Classes ...............................................................44

Classes LowArray and LowArrayApp .................................................46

Class Interfaces ..........................................................................................46

Not So Convenient ...........................................................................47

Who’s Responsible for What? ..........................................................48

The highArray.java Example ..........................................................48

The User’s Life Made Easier ..............................................................52

Abstraction .......................................................................................52

The Ordered Workshop Applet .................................................................52

Linear Search ....................................................................................53

Binary Search ....................................................................................54

Java Code for an Ordered Array ................................................................56

Binary Search with the find() Method ...........................................56

The OrdArray Class ..........................................................................58

Advantages of Ordered Arrays .........................................................61

Logarithms .................................................................................................62

The Equation ....................................................................................63

The Opposite of Raising Two to a Power .........................................64

00 0672324539 fm 10/10/02 9:13 AM Page vi vi

Data Structures & Algorithms in Java, Second Edition Storing Objects ..........................................................................................64

The Person Class ..............................................................................65

The classDataArray.java Program ................................................65

Big O Notation .........................................................................................70

Insertion in an Unordered Array: Constant ....................................70

Linear Search: Proportional to N .....................................................70

Binary Search: Proportional to log(N) .............................................71

Don’t Need the Constant .................................................................71

Why Not Use Arrays for Everything? ........................................................72

Summary ....................................................................................................73

Questions ...................................................................................................74

Experiments ...............................................................................................75

Programming Projects ...............................................................................76

**3**

**Simple Sorting **

**77**

How Would You Do It? .............................................................................78

Bubble Sort .................................................................................................79

Bubble Sort on the Baseball Players .................................................79

The BubbleSort Workshop Applet ....................................................81

Java Code for a Bubble Sort .............................................................85

Invariants ..........................................................................................88

Efficiency of the Bubble Sort ...........................................................88

Selection Sort .............................................................................................89

Selection Sort on the Baseball Players .............................................89

The SelectSort Workshop Applet .....................................................90

Java Code for Selection Sort .............................................................92

Invariant ...........................................................................................95

Efficiency of the Selection Sort ........................................................95

Insertion Sort .............................................................................................95

Insertion Sort on the Baseball Players .............................................95

The InsertSort Workshop Applet .....................................................97

Java Code for Insertion Sort .............................................................99

Invariants in the Insertion Sort .....................................................103

Efficiency of the Insertion Sort ......................................................103

Sorting Objects ........................................................................................103

Java Code for Sorting Objects ........................................................104

Lexicographical Comparisons ........................................................107

Stability ...........................................................................................107

Comparing the Simple Sorts ...................................................................108

Summary ..................................................................................................108

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vii

Questions .................................................................................................109

Experiments .............................................................................................111

Programming Projects .............................................................................112

**4**

**Stacks and Queues **

**115**

A Different Kind of Structure ..................................................................115

Programmer’s Tools ........................................................................115

Restricted Access .............................................................................116

More Abstract .................................................................................116

Stacks .......................................................................................................116

The Postal Analogy .........................................................................117

The Stack Workshop Applet ...........................................................118

Java Code for a Stack ......................................................................120

Stack Example 1: Reversing a Word ...............................................124

Stack Example 2: Delimiter Matching ...........................................127

Efficiency of Stacks .........................................................................132

Queues .....................................................................................................132

The Queue Workshop Applet .........................................................133

A Circular Queue ............................................................................136

Java Code for a Queue ...................................................................137

Efficiency of Queues .......................................................................142

Deques ............................................................................................143

Priority Queues ........................................................................................143

The PriorityQ Workshop Applet ....................................................144

Java Code for a Priority Queue ......................................................147

Efficiency of Priority Queues .........................................................149

Parsing Arithmetic Expressions ...............................................................149

Postfix Notation .............................................................................150

Translating Infix to Postfix ............................................................151

Evaluating Postfix Expressions .......................................................167

Summary ..................................................................................................173

Questions .................................................................................................174

Experiments .............................................................................................176

Programming Projects .............................................................................176

**5**

**Linked Lists **

**179**

Links .........................................................................................................179

References and Basic Types ............................................................180

Relationship, Not Position .............................................................182

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Data Structures & Algorithms in Java, Second Edition The LinkList Workshop Applet ...............................................................183

The Insert Button ...........................................................................183

The Find Button .............................................................................184

The Delete Button ..........................................................................184

A Simple Linked List ................................................................................185

The Link Class ................................................................................185

The LinkList Class ........................................................................186

The insertFirst() Method ...........................................................187

The deleteFirst() Method ...........................................................188

The displayList() Method ...........................................................189

The linkList.java Program ..........................................................190

Finding and Deleting Specified Links .....................................................193

The find() Method ........................................................................196

The delete() Method ....................................................................196

Other Methods ...............................................................................197

Double-Ended Lists ..................................................................................198

Linked-List Efficiency ..............................................................................202

Abstract Data Types .................................................................................202

A Stack Implemented by a Linked List ..........................................203

A Queue Implemented by a Linked List ........................................206

Data Types and Abstraction ...........................................................210

ADT Lists ........................................................................................211

ADTs as a Design Tool ....................................................................212

Sorted Lists ...............................................................................................212

Java Code to Insert an Item in a Sorted List .................................213

The sortedList.java Program ......................................................215

Efficiency of Sorted Linked Lists ....................................................218

List Insertion Sort ...........................................................................218

Doubly Linked Lists .................................................................................221

Traversal ..........................................................................................222

Insertion .........................................................................................223

Deletion ..........................................................................................225

The doublyLinked.java Program ..................................................226

Doubly Linked List as Basis for Deques .........................................231

Iterators ....................................................................................................231

A Reference in the List Itself? ........................................................232

An Iterator Class .............................................................................232

Additional Iterator Features ...........................................................233

Iterator Methods .............................................................................234

The interIterator.java Program ................................................235

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ix

Where Does the Iterator Point? .....................................................242

The atEnd() Method ......................................................................242

Iterative Operations ........................................................................243

Other Methods ...............................................................................244

Summary ..................................................................................................244

Questions .................................................................................................245

Experiments .............................................................................................247

Programming Projects .............................................................................247

**6**

**Recursion 251**

Triangular Numbers .................................................................................251

Finding the nth Term Using a Loop ..............................................252

Finding the nth Term Using Recursion ..........................................253

The triangle.java Program ..........................................................255

What’s Really Happening? .............................................................257

Characteristics of Recursive Methods ............................................259

Is Recursion Efficient? ....................................................................259

Mathematical Induction ................................................................259

Factorials ..................................................................................................260

Anagrams .................................................................................................262

A Recursive Binary Search .......................................................................268

Recursion Replaces the Loop .........................................................268

Divide-and-Conquer Algorithms ...................................................272

The Towers of Hanoi ...............................................................................273

The Towers Workshop Applet ........................................................274

Moving Subtrees .............................................................................275

The Recursive Algorithm ................................................................276

The towers.java Program .............................................................277

mergesort .................................................................................................279

Merging Two Sorted Arrays ............................................................280

Sorting by Merging .........................................................................283

The MergeSort Workshop Applet ...................................................285

The mergeSort.java Program ........................................................287

Efficiency of the mergesort ............................................................291

Eliminating Recursion .............................................................................294

Recursion and Stacks ......................................................................294

Simulating a Recursive Method .....................................................294

What Does This Prove? ..................................................................301

Some Interesting Recursive Applications ................................................303

Raising a Number to a Power .........................................................303

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Data Structures & Algorithms in Java, Second Edition The Knapsack Problem ...................................................................305

Combinations: Picking a Team ......................................................306

Summary ..................................................................................................308

Questions .................................................................................................310

Experiments .............................................................................................312

Programming Projects .............................................................................312

**7**

**Advanced Sorting **

**315**

Shellsort ...................................................................................................315

Insertion Sort: Too Many Copies ...................................................316

N-Sorting ........................................................................................316

Diminishing Gaps ..........................................................................317

The Shellsort Workshop Applet .....................................................319

Java Code for the Shellsort ............................................................321

Other Interval Sequences ...............................................................324

Efficiency of the Shellsort ..............................................................324

Partitioning ..............................................................................................325

The Partition Workshop Applet .....................................................325

The partition.java Program ........................................................327

The Partition Algorithm .................................................................330

Efficiency of the Partition Algorithm ............................................332

Quicksort .................................................................................................333

The Quicksort Algorithm ...............................................................333

Choosing a Pivot Value ..................................................................335

The QuickSort1 Workshop Applet .................................................340

Degenerates to O(N2) Performance ...............................................344

Median-of-Three Partitioning ........................................................345

Handling Small Partitions ..............................................................350

Removing Recursion ......................................................................354

Efficiency of Quicksort ...................................................................355

Radix Sort .................................................................................................357

Algorithm for the Radix Sort .........................................................358

Designing a Program ......................................................................358

Efficiency of the Radix Sort ...........................................................359

Summary ..................................................................................................359

Questions .................................................................................................361

Experiments .............................................................................................363

Programming Projects .............................................................................363

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**8**

**Binary Trees **

**365**

Why Use Binary Trees? ............................................................................365

Slow Insertion in an Ordered Array ...............................................365

Slow Searching in a Linked List .....................................................366

Trees to the Rescue .........................................................................366

What Is a Tree? ...............................................................................366

Tree Terminology .....................................................................................367

Path .................................................................................................368

Root ................................................................................................368

Parent ..............................................................................................369

Child ...............................................................................................369

Leaf .................................................................................................369

Subtree ............................................................................................369

Visiting ...........................................................................................369

Traversing .......................................................................................369

Levels ..............................................................................................369

Keys .................................................................................................369

Binary Trees ....................................................................................370

An Analogy ..............................................................................................370

How Do Binary Search Trees Work? ........................................................371

The Binary Tree Workshop Applet .................................................371

Representing the Tree in Java Code ...............................................373

Finding a Node ........................................................................................376

Using the Workshop Applet to Find a Node .................................376

Java Code for Finding a Node ........................................................377

Tree Efficiency ................................................................................378

Inserting a Node ......................................................................................378

Using the Workshop Applet to Insert a Node ...............................379

Java Code for Inserting a Node ......................................................379

Traversing the Tree ..................................................................................381

Inorder Traversal ............................................................................381

Java Code for Traversing ................................................................382

Traversing a Three-Node Tree ........................................................382

Traversing with the Workshop Applet ...........................................384

Preorder and Postorder Traversals ..................................................385

Finding Maximum and Minimum Values ..............................................388

Deleting a Node .......................................................................................389

Case 1: The Node to Be Deleted Has No Children ........................389

Case 2: The Node to Be Deleted Has One Child ...........................391

Case 3: The Node to Be Deleted Has Two Children ......................393

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Data Structures & Algorithms in Java, Second Edition The Efficiency of Binary Trees .................................................................401

Trees Represented as Arrays .....................................................................403

Duplicate Keys .........................................................................................404

The Complete tree.java Program .........................................................405

The Huffman Code ..................................................................................415

Character Codes .............................................................................415

Decoding with the Huffman Tree ..................................................417

Creating the Huffman Tree ............................................................418

Coding the Message .......................................................................420

Creating the Huffman Code ..........................................................421

Summary ..................................................................................................422

Questions .................................................................................................423

Experiments .............................................................................................425

Programming Projects .............................................................................425

**9**

**Red-Black Trees **

**429**

Our Approach to the Discussion .............................................................429

Conceptual .....................................................................................430

Top-Down Insertion .......................................................................430

Balanced and Unbalanced Trees ..............................................................430

Degenerates to O(N) .......................................................................431

Balance to the Rescue .....................................................................432

Red-Black Tree Characteristics .......................................................432

Fixing Rule Violations ....................................................................434

Using the RBTree Workshop Applet ........................................................434

Clicking on a Node ........................................................................435

The Start Button .............................................................................435

The Ins Button ...............................................................................435

The Del Button ...............................................................................436

The Flip Button ..............................................................................436

The RoL Button ..............................................................................436

The RoR Button ..............................................................................436

The R/B Button ...............................................................................436

Text Messages .................................................................................437

Where’s the Find Button? ..............................................................437

Experimenting with the Workshop Applet .............................................437

Experiment 1: Inserting Two Red Nodes .......................................437

Experiment 2: Rotations ................................................................438

Experiment 3: Color Flips ..............................................................439

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Experiment 4: An Unbalanced Tree ...............................................439

More Experiments ..........................................................................440

The Red-Black Rules and Balanced Trees .......................................440

Null Children .................................................................................441

Rotations ..................................................................................................441

Simple Rotations ............................................................................442

The Weird Crossover Node ............................................................442

Subtrees on the Move ....................................................................444

Human Beings Versus Computers ..................................................445

Inserting a New Node ..............................................................................445

Preview of the Insertion Process ....................................................446

Color Flips on the Way Down .......................................................446

Rotations After the Node Is Inserted .............................................448

Rotations on the Way Down .........................................................454

Deletion ...................................................................................................457

The Efficiency of Red-Black Trees ............................................................457

Red-Black Tree Implementation ..............................................................458

Other Balanced Trees ...............................................................................458

Summary ..................................................................................................459

Questions .................................................................................................460

Experiments .............................................................................................462

**10**

**2-3-4 Trees and External Storage **

**463**

Introduction to 2-3-4 Trees .....................................................................463

What’s in a Name? .........................................................................464

2-3-4 Tree Organization .................................................................465

Searching a 2-3-4 Tree ....................................................................466

Insertion .........................................................................................466

Node Splits .....................................................................................467

Splitting the Root ...........................................................................468

Splitting on the Way Down ...........................................................469

The Tree234 Workshop Applet ................................................................470

The Fill Button ...............................................................................471

The Find Button .............................................................................471

The Ins Button ...............................................................................472

The Zoom Button ...........................................................................472

Viewing Different Nodes ................................................................473

Experiments ....................................................................................474

Java Code for a 2-3-4 Tree .......................................................................475

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Data Structures & Algorithms in Java, Second Edition The DataItem Class ........................................................................475

The Node Class ................................................................................475

The Tree234 Class ..........................................................................476

The Tree234App Class .....................................................................477

The Complete tree234.java Program ...........................................478

2-3-4 Trees and Red-Black Trees ..............................................................486

Transformation from 2-3-4 to Red-Black .......................................486

Operational Equivalence ................................................................488

Efficiency of 2-3-4 Trees ..........................................................................491

Speed ...............................................................................................491

Storage Requirements .....................................................................491

2-3 Trees ...................................................................................................492

Node Splits .....................................................................................492

Implementation .............................................................................494

External Storage .......................................................................................496

Accessing External Data .................................................................496

Sequential Ordering .......................................................................499

B-Trees .............................................................................................500

Indexing .........................................................................................506

Complex Search Criteria ................................................................509

Sorting External Files .....................................................................509

Summary ..................................................................................................513

Questions .................................................................................................514

Experiments .............................................................................................516

Programming Projects .............................................................................516

**11**

**Hash Tables **

**519**

Introduction to Hashing .........................................................................520

Employee Numbers as Keys ...........................................................520

A Dictionary ...................................................................................521

Hashing ..........................................................................................525

Collisions ........................................................................................527

Open Addressing .....................................................................................528

Linear Probing ................................................................................528

Java Code for a Linear Probe Hash Table .......................................533

Quadratic Probing ..........................................................................542

Double Hashing ..............................................................................544

Separate Chaining ...................................................................................552

The HashChain Workshop Applet .................................................552

Java Code for Separate Chaining ...................................................555

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xv

Hash Functions ........................................................................................561

Quick Computation .......................................................................561

Random Keys ..................................................................................562

Non-Random Keys .........................................................................562

Hashing Strings ..............................................................................563

Folding ............................................................................................566

Hashing Efficiency ...................................................................................566

Open Addressing ............................................................................566

Separate Chaining ..........................................................................568

Open Addressing Versus Separate Chaining ..................................570

Hashing and External Storage .................................................................571

Table of File Pointers ......................................................................571

Non-Full Blocks ..............................................................................571

Full Blocks ......................................................................................572

Summary ..................................................................................................573

Questions .................................................................................................574

Experiments .............................................................................................576

Programming Projects .............................................................................577

**12**

**Heaps 579**

Introduction to Heaps .............................................................................580

Priority Queues, Heaps, and ADTs .................................................581

Weakly Ordered ..............................................................................582

Removal ..........................................................................................583

Insertion .........................................................................................585

Not Really Swapped .......................................................................586

The Heap Workshop Applet ....................................................................587

The Fill Button ...............................................................................587

The Change Button ........................................................................588

The Remove Button .......................................................................588

The Insert Button ...........................................................................588

Java Code for Heaps ................................................................................588

Insertion .........................................................................................589

Removal ..........................................................................................590

Key Change ....................................................................................591

The Array Size .................................................................................592

The heap.java Program .................................................................592

Expanding the Heap Array .............................................................599

Efficiency of Heap Operations .......................................................599

00 0672324539 fm 10/10/02 9:13 AM Page xvi xvi

Data Structures & Algorithms in Java, Second Edition A Tree-based Heap ...................................................................................600

Heapsort ...................................................................................................601

Trickling Down in Place .................................................................602

Using the Same Array .....................................................................604

The heapSort.java Program ..........................................................605

The Efficiency of Heapsort .............................................................610

Summary ..................................................................................................610

Questions .................................................................................................611

Experiments .............................................................................................612

Programming Projects .............................................................................612

**13**

**Graphs 615**

Introduction to Graphs ...........................................................................615

Definitions ......................................................................................616

Historical Note ...............................................................................618

Representing a Graph in a Program ..............................................619

Adding Vertices and Edges to a Graph ..........................................622

The Graph Class ..............................................................................622

Searches ....................................................................................................623

Depth-First Search ..........................................................................625

Breadth-First Search ........................................................................636

Minimum Spanning Trees .......................................................................643

GraphN Workshop Applet .............................................................644

Java Code for the Minimum Spanning Tree ..................................644

The mst.java Program ...................................................................645

Topological Sorting with Directed Graphs ..............................................649

An Example: Course Prerequisites .................................................649

Directed Graphs .............................................................................650

Topological Sorting ........................................................................651

The GraphD Workshop Applet ......................................................652

Cycles and Trees .............................................................................653

Java Code ........................................................................................654

Connectivity in Directed Graphs ............................................................661

The Connectivity Table ..................................................................662

Warshall’s Algorithm ......................................................................662

Implementation of Warshall’s Algorithm ......................................664

Summary ..................................................................................................665

Questions .................................................................................................665

Experiments .............................................................................................667

Programming Projects .............................................................................667

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xvii

**14**

**Weighted Graphs **

**669**

Minimum Spanning Tree with Weighted Graphs ..................................669

An Example: Cable TV in the Jungle .............................................670

The GraphW Workshop Applet .....................................................670

Send Out the Surveyors ..................................................................672

Creating the Algorithm ..................................................................676

Java Code ........................................................................................678

The mstw.java Program .................................................................681

The Shortest-Path Problem ......................................................................687

The Railroad Line ...........................................................................687

Dijkstra’s Algorithm .......................................................................689

Agents and Train Rides ...................................................................689

Using the GraphDW Workshop Applet .........................................694

Java Code ........................................................................................698

The path.java Program .................................................................703

The All-Pairs Shortest-Path Problem .......................................................708

Efficiency .................................................................................................710

Intractable Problems ................................................................................710

The Knight’s Tour ...........................................................................711

The Traveling Salesman Problem ...................................................711

Hamiltonian Cycles ........................................................................712

Summary ..................................................................................................713

Questions .................................................................................................713

Experiments .............................................................................................715

Programming Projects .............................................................................715

**15**

**When to Use What **

**717**

General-Purpose Data Structures .............................................................717

Speed and Algorithms ....................................................................718

Libraries ..........................................................................................719

Arrays ..............................................................................................720

Linked Lists .....................................................................................720

Binary Search Trees .........................................................................720

Balanced Trees ................................................................................721

Hash Tables .....................................................................................721

Comparing the General-Purpose Storage Structures .....................722

Special-Purpose Data Structures ..............................................................722

Stack ................................................................................................723

Queue .............................................................................................723

00 0672324539 fm 10/10/02 9:13 AM Page xviii xviii

Data Structures & Algorithms in Java, Second Edition Priority Queue ................................................................................723

Comparison of Special-Purpose Structures ....................................724

Sorting ......................................................................................................724

Graphs ......................................................................................................725

External Storage .......................................................................................725

Sequential Storage ..........................................................................726

Indexed Files ...................................................................................726

B-trees .............................................................................................726

Hashing ..........................................................................................727

Virtual Memory ..............................................................................727

Onward ....................................................................................................728

**Appendixes**

**A**

**Running the Workshop Applets and Example Programs** **729**

The Workshop Applets ............................................................................729

The Example Programs ............................................................................730

The Sun Microsystem’s Software Development Kit ................................730

Command-line Programs ...............................................................731

Setting the Path ..............................................................................731

Viewing the Workshop Applets .....................................................731

Operating the Workshop Applets ..................................................732

Running the Example Programs ....................................................732

Compiling the Example Programs .................................................733

Editing the Source Code .................................................................733

Terminating the Example Programs ..............................................733

Multiple Class Files ..................................................................................733

Other Development Systems ...................................................................734

**B**

**Further Reading **

**735**

Data Structures and Algorithms ..............................................................735

Object-Oriented Programming Languages ..............................................736

Object-Oriented Design (OOD) and Software Engineering ....................736

**C**

**Answers to Questions **

**739**

Chapter 1, Overview ................................................................................739

Answers to Questions .....................................................................739

Chapter 2, Arrays .....................................................................................739

Answers to Questions .....................................................................739

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xix

Chapter 3, Simple Sorting .......................................................................740

Answers to Questions .....................................................................740

Chapter 4, Stacks and Queues .................................................................741

Answers to Questions .....................................................................741

Chapter 5, Linked Lists ...........................................................................741

Answers to Questions .....................................................................741

Chapter 6, Recursion ...............................................................................742

Answers to Questions .....................................................................742

Chapter 7, Advanced Sorting ..................................................................743

Answers to Questions .....................................................................743

Chapter 8, Binary Trees ...........................................................................743

Answers to Questions .....................................................................743

Chapter 9, Red-Black Trees ......................................................................744

Answers to Questions .....................................................................744

Chapter 10, 2-3-4 Trees and External Storage .........................................745

Answers to Questions .....................................................................745

Chapter 11, Hash Tables ..........................................................................745

Answers to Questions .....................................................................745

Chapter 12, Heaps ...................................................................................746

Answers to Questions .....................................................................746

Chapter 13, Graphs .................................................................................746

Answers to Questions .....................................................................746

Chapter 14, Weighted Graphs .................................................................747

Answers to Questions .....................................................................747

00 0672324539 fm 10/10/02 3:17 PM Page xx

About the Author

**Robert Lafore **has degrees in Electrical Engineering and Mathematics, has worked as a systems analyst for the Lawrence Berkeley Laboratory, founded his own software company, and is a best-selling writer in the field of computer programming.

Some of his current titles are

00 0672324539 fm 10/10/02 9:13 AM Page xxi Dedication

00 0672324539 fm 10/10/02 9:13 AM Page xxii Acknowledgments to the First Edition

My gratitude for the following people (and many others) cannot be fully expressed in this short acknowledgment. As always, Mitch Waite had the Java thing figured out before anyone else. He also let me bounce the applets off him until they did the job, and extracted the overall form of the project from a miasma of speculation. My editor, Kurt Stephan, found great reviewers, made sure everyone was on the same page, kept the ball rolling, and gently but firmly ensured that I did what I was supposed to do. Harry Henderson provided a skilled appraisal of the first draft, along with many valuable suggestions. Richard S. Wright, Jr., as technical editor, corrected numerous problems with his keen eye for detail. Jaime Niño, Ph.D., of the University of New Orleans, attempted to save me from myself and occasionally succeeded, but should bear no responsibility for my approach or coding details. Susan Walton has been a staunch and much-appreciated supporter in helping to convey the essence of the project to the non-technical. Carmela Carvajal was invaluable in extending our contacts with the academic world. Dan Scherf not only put the CD-ROM together, but was tireless in keeping me up to date on rapidly evolving software changes.

Finally, Cecile Kaufman ably shepherded the book through its transition from the editing to the production process.

Acknowledgments to the Second Edition

My thanks to the following people at Sams Publishing for their competence, effort, and patience in the development of this second edition. Acquisitions Editor Carol Ackerman and Development Editor Songlin Qiu ably guided this edition through the complex production process. Project Editor Matt Purcell corrected a semi-infinite number of grammatical errors and made sure everything made sense. Tech Editor Mike Kopak reviewed the programs and saved me from several problems. Last but not least, Dan Scherf, an old friend from a previous era, provides skilled management of my code and applets on the Sams Web site.

00 0672324539 fm 8/28/03 9:35 AM Page xxiii We Want to Hear from You!

As the reader of this book,

As an executive editor for Sams Publishing, I welcome your comments. You can email or write me directly to let me know what you did or didn’t like about this book—as well as what we can do to make our books better.

Please note that I cannot help you with technical problems related to the

When you write, please be sure to include this book’s title and author as well as your name, email address, and phone number. I will carefully review your comments and share them with the author and editors who worked on the book.

Email:

feedback@samspublishing.com

Mail:

Michael Stephens

Executive Editor

Sams Publishing

800 East 96th Street

Indianapolis, IN 46240 USA

For more information about this book or another Sams Publishing title, visit our Web site at www.samspublishing.com. Type the ISBN (excluding hyphens) or the title of a book in the Search field to find the page you’re looking for.

00 0672324539 fm 10/10/02 9:13 AM Page xxiv

01 0672324539 intro 10/10/02 9:08 AM Page 1

Introduction

This introduction tells you briefly

• What’s new in the Second Edition

• What this book is about

• Why it’s different

• Who might want to read it

• What you need to know before you read it

• The software and equipment you need to use it

• How this book is organized

**What’s New in the Second Edition**

This second edition of

**Additional Topics**

We’ve added a variety of interesting new topics to the book. Many provide a basis for programming projects. These new topics include

• Depth-first-search and game simulations

• The Josephus problem

• Huffman codes for data compression

• The Traveling Salesman problem

• Hamiltonian cycles

• The Knight’s Tour puzzle

• Floyd’s algorithm

• Warshall’s algorithm

• 2-3 trees

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2

Data Structures & Algorithms in Java, Second Edition

• The knapsack problem

• Listing N things taken K at a time

• Folding-digits hash functions

• The radix sort

**End-of-Chapter Questions**

Short questions covering the key points of each chapter are included at the end of each chapter. The answers can be found in Appendix C, “Answers to Questions.”

These questions are intended as a self-test for readers, to ensure that they have understood the material.

**Experiments**

We include some suggested activities for the reader. These experiments often involve using the Workshop applets or example programs to examine certain features of an algorithm’s operation, but some are pencil-and-paper or “thought experiments.”

**Programming Projects**

Most importantly, we have included at the end of each chapter a number (usually five) of challenging programming projects. They cover a range of difficulty. The easiest are simple variations on the example programs. The most challenging are implementations of topics discussed in the text but for which there are no example programs. Solutions to the Programming Projects are not provided in this book, but see the adjacent note.

**NOTE**

It is expected that the programming projects will be useful for instructors looking for class assignments. To this end, qualified instructors can obtain suggested solutions to the programming projects in the form of source code and executable code. Contact the Sams Web site for information on Instructors Programs.

**What This Book Is About**

This book is about data structures and algorithms as used in computer programming.

Data structures are ways in which data is arranged in your computer’s memory (or stored on disk). Algorithms are the procedures a software program uses to manipulate the data in these structures.

01 0672324539 intro 10/10/02 9:08 AM Page 3

Introduction

3

Almost every computer program, even a simple one, uses data structures and algorithms. For example, consider a program that prints address labels. The program might use an array containing the addresses to be printed and a simple for loop to step through the array, printing each address.

The array in this example is a data structure, and the for loop, used for sequential access to the array, executes a simple algorithm. For uncomplicated programs with small amounts of data, such a simple approach might be all you need. However, for programs that handle even moderately large amounts of data, or which solve problems that are slightly out of the ordinary, more sophisticated techniques are necessary. Simply knowing the syntax of a computer language such as Java or C++ isn’t enough.

This book is about what you need to know

**What’s Different About This Book**

There are dozens of books on data structures and algorithms. What’s different about this one? Three things:

• Our primary goal in writing this book is to make the topics we cover easy to understand.

• Demonstration programs called

• The example code is written in Java, which is easier to understand than C, C++, or Pascal, the languages traditionally used to demonstrate computer science topics.

Let’s look at these features in more detail.

**Easy to Understand**

Typical computer science textbooks are full of theory, mathematical formulas, and abstruse examples of computer code. This book, on the other hand, concentrates on simple explanations of techniques that can be applied to real-world problems. We avoid complex proofs and heavy math. There are lots of figures to augment the text.

Many books on data structures and algorithms include considerable material on software engineering. Software engineering is a body of study concerned with designing and implementing large and complex software projects.

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4

Data Structures & Algorithms in Java, Second Edition However, it’s our belief that data structures and algorithms are complicated enough without involving this additional discipline, so we have deliberately de-emphasized software engineering in this book. (We’ll discuss the relationship of data structures and algorithms to software engineering in Chapter 1, “Overview.”) Of course, we do use an object-oriented approach, and we discuss various aspects of object-oriented design as we go along, including a mini-tutorial on OOP in Chapter 1. Our primary emphasis, however, is on the data structures and algorithms themselves.

**Workshop Applets**

From the Sams Web site you can download demonstration programs, in the form of Java applets, that cover the topics we discuss. These applets, which we call

For example, in one Workshop applet, each time you push a button, a bar chart shows you one step in the process of sorting the bars into ascending order. The values of variables used in the sorting algorithm are also shown, so you can see exactly how the computer code works when executing the algorithm. Text displayed in the picture explains what’s happening.

Another applet models a binary tree. Arrows move up and down the tree, so you can follow the steps involved in inserting or deleting a node from the tree. There are more than 20 Workshop applets, at least one for each of the major topics in the book.

These Workshop applets make it far more obvious what a data structure really looks like, or what an algorithm is supposed to do, than a text description ever could. Of course, we provide a text description as well. The combination of Workshop applets, clear text, and illustrations should make things easy.

These Workshop applets are standalone graphics-based programs. You can use them as a learning tool that augments the material in the book. Note that they’re not the same as the example code found in the text of the book, which we’ll discuss next.

**NOTE**

The Workshop applets, in the form of Java .class files, are available on the Sams Web site at http://www.samspublishing.com/

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Introduction

5

**Java Example Code**

The Java language is easier to understand (and write) than languages such as C and C++. The biggest reason for this is that Java doesn’t use pointers. Some people are surprised that pointers aren’t necessary for the creation of complex data structures and algorithms. In fact, eliminating pointers makes such code not only easier to write and to understand, but more secure and less prone to errors as well.

Java is a modern object-oriented language, which means we can use an object-oriented approach for the programming examples. This is important, because object-oriented programming (OOP) offers compelling advantages over the old-fashioned procedural approach, and is quickly supplanting it for serious program development.

Don’t be alarmed if you aren’t familiar with OOP. It’s not that hard to understand, especially in a pointer-free environment such as Java. We’ll explain the basics of OOP in Chapter 1.

**NOTE**

Like the Workshop applets, the example programs (both source and executable files) can be downloaded from the Sams Web site.

**Who This Book Is For**

This book can be used as a text in a Data Structures and Algorithms course, typically taught in the second year of a computer science curriculum. However, it is also designed for professional programmers and for anyone else who needs to take the next step up from merely knowing a programming language. Because it’s easy to understand, it is also appropriate as a supplemental text to a more formal course.

**What You Need to Know Before You Read This Book**

The only prerequisite for using this book is a knowledge of some programming language.

Although the example code is written in Java, you don’t need to know Java to follow what’s happening. Java is not hard to understand, and we’ve tried to keep the syntax as general as possible, avoiding baroque or Java-specific constructions whenever possible.

Of course, it won’t hurt if you’re already familiar with Java. Knowing C++ is essentially just as good, because Java syntax is based so closely on C++. The differences are minor as they apply to our example programs (except for the welcome elimination of pointers), and we’ll discuss them in Chapter 1.

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6

Data Structures & Algorithms in Java, Second Edition **The Software You Need to Use This Book**

To run the Workshop applets, you need a Web browser such as Microsoft Internet Explorer or Netscape Communicator. You can also use an applet viewer utility.

Applet viewers are available with various Java development systems, including the free system from Sun Microsystems, which we’ll discuss in Appendix A.

To run the example programs, you can use the MS-DOS utility in Microsoft Windows (called MS-DOS Prompt) or a similar text-oriented environment.

If you want to modify the source code for the example programs or write your own programs, you’ll need a Java development system. Such systems are available commercially, or you can download an excellent basic system from Sun Microsystems, as described in Appendix A.

**How This Book Is Organized**

This section is intended for teachers and others who want a quick overview of the contents of the book. It assumes you’re already familiar with the topics and terms involved in a study of data structures and algorithms.

The first two chapters are intended to ease the reader into data structures and algorithms as painlessly as possible.

Chapter 1, “Overview,” presents an overview of the topics to be discussed and introduces a small number of terms that will be needed later on. For readers unfamiliar with object-oriented programming, it summarizes those aspects of this discipline that will be needed in the balance of the book, and for programmers who know C++

but not Java, the key differences between these languages are reviewed.

Chapter 2, “Arrays,” focuses on arrays. However, there are two subtexts: the use of classes to encapsulate data storage structures and the class interface. Searching, insertion, and deletion in arrays and ordered arrays are covered. Linear searching and binary searching are explained. Workshop applets demonstrate these algorithms with unordered and ordered arrays.

In Chapter 3, “Simple Sorting,” we introduce three simple (but slow) sorting techniques: the bubble sort, selection sort, and insertion sort. Each is demonstrated by a Workshop applet.

Chapter 4, “Stacks and Queues,” covers three data structures that can be thought of as Abstract Data Types (ADTs): the stack, queue, and priority queue. These structures reappear later in the book, embedded in various algorithms. Each is demonstrated by a Workshop applet. The concept of ADTs is discussed.

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Introduction

7

Chapter 5, “Linked Lists,” introduces linked lists, including doubly linked lists and double-ended lists. The use of references as “painless pointers” in Java is explained. A Workshop applet shows how insertion, searching, and deletion are carried out.

In Chapter 6, “Recursion,” we explore recursion, one of the few chapter topics that is not a data structure. Many examples of recursion are given, including the Towers of Hanoi puzzle and the mergesort, which are demonstrated by Workshop applets.

Chapter 7, “Advanced Sorting,” delves into some advanced sorting techniques: Shellsort and quicksort. Workshop applets demonstrate Shellsort, partitioning (the basis of quicksort), and two flavors of quicksort.

In Chapter 8, “Binary Trees,” we begin our exploration of trees. This chapter covers the simplest popular tree structure: unbalanced binary search trees. A Workshop applet demonstrates insertion, deletion, and traversal of such trees.

Chapter 9, “Red-Black Trees,” explains red-black trees, one of the most efficient balanced trees. The Workshop applet demonstrates the rotations and color switches necessary to balance the tree.

In Chapter 10, “2-3-4 Trees and External Storage,” we cover 2-3-4 trees as an example of multiway trees. A Workshop applet shows how they work. We also discuss 2-3

trees and the relationship of 2-3-4 trees to B-trees, which are useful in storing external (disk) files.

Chapter 11, “Hash Tables,” moves into a new field, hash tables. Workshop applets demonstrate several approaches: linear and quadratic probing, double hashing, and separate chaining. The hash-table approach to organizing external files is discussed.

In Chapter 12, “Heaps,” we discuss the heap, a specialized tree used as an efficient implementation of a priority queue.

Chapters 13, “Graphs,” and 14, “Weighted Graphs,” deal with graphs, the first with unweighted graphs and simple searching algorithms, and the second with weighted graphs and more complex algorithms involving the minimum spanning trees and shortest paths.

In Chapter 15, “When to Use What,” we summarize the various data structures described in earlier chapters, with special attention to which structure is appropriate in a given situation.

Appendix A, “Running the Workshop Applets and Example Programs,” provides details on how to use these two kinds of software. It also tells how to use the Software Development Kit from Sun Microsystems, which can be used to modify the example programs and develop your own programs, and to run the applets and example programs.

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8

Data Structures & Algorithms in Java, Second Edition Appendix B, “Further Reading,” describes some books appropriate for further reading on data structures and other related topics.

Appendix C, “Answers to Questions,” contains the answers to the end-of-chapter questions in the text.

**Enjoy Yourself! **

We hope we’ve made the learning process as painless as possible. Ideally, it should even be fun. Let us know if you think we’ve succeeded in reaching this ideal, or if not, where you think improvements might be made.

02 0672324539 CH01 10/10/02 9:10 AM Page 9

**1**

**IN THIS CHAPTER**

• What Are Data Structures and

Overview

Algorithms Good For?

• Overview of Data Structures

• Overview of Algorithms

As you start this book, you may have some questions:

• Some Definitions

• What are data structures and algorithms?

• Object-Oriented

• What good will it do me to know about them?

Programming

• Why can’t I just use arrays and for loops to handle

• Software Engineering

my data?

• Java for C++ Programmers

• When does it make sense to apply what I learn here?

• Java Library Data Structures

This chapter attempts to answer these questions. We’ll also introduce some terms you’ll need to know and generally set the stage for the more detailed chapters to follow.

Next, for those of you who haven’t yet been exposed to an object-oriented language, we’ll briefly explain enough about OOP to get you started. Finally, for C++ programmers who don’t know Java we’ll point out some of the differences between these languages.

**What Are Data Structures and**

**Algorithms Good For? **

The subject of this book is data structures and algorithms.

A

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10

**CHAPTER 1**

Overview

What sorts of problems can you solve with a knowledge of these topics? As a rough approximation, we might divide the situations in which they’re useful into three categories:

• Real-world data storage

• Programmer’s tools

• Modeling

These are not hard-and-fast categories, but they may help give you a feeling for the usefulness of this book’s subject matter. Let’s look at them in more detail.

**Real-World Data Storage**

Many of the structures and techniques we’ll discuss are concerned with how to handle real-world data storage. By real-world data, we mean data that describes physical entities external to the computer. As some examples, a personnel record describes an actual human being, an inventory record describes an existing car part or grocery item, and a financial transaction record describes, say, an actual check written to pay the electric bill.

A non-computer example of real-world data storage is a stack of 3-by-5 index cards.

These cards can be used for a variety of purposes. If each card holds a person’s name, address, and phone number, the result is an address book. If each card holds the name, location, and value of a household possession, the result is a home inventory.

Of course, index cards are not exactly state-of-the-art. Almost anything that was once done with index cards can now be done with a computer. Suppose you want to update your old index-card system to a computer program. You might find yourself with questions like these:

• How would you store the data in your computer’s memory?

• Would your method work for a hundred file cards? A thousand? A million?

• Would your method permit quick insertion of new cards and deletion of old ones?

• Would it allow for fast searching for a specified card?

• Suppose you wanted to arrange the cards in alphabetical order. How would you sort them?

In this book, we will be discussing data structures that might be used in ways similar to a stack of index cards.

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Overview of Data Structures

11

Of course, most programs are more complex than index cards. Imagine the database the Department of Motor Vehicles (or whatever it’s called in your state) uses to keep track of drivers’ licenses, or an airline reservations system that stores passenger and flight information. Such systems may include many data structures. Designing such complex systems requires the application of software engineering techniques, which we’ll mention toward the end of this chapter.

**Programmer’s Tools**

Not all data storage structures are used to store real-world data. Typically, real-world data is accessed more or less directly by a program’s user. Some data storage structures, however, are not meant to be accessed by the user, but by the program itself. A programmer uses such structures as tools to facilitate some other operation. Stacks, queues, and priority queues are often used in this way. We’ll see examples as we go along.

**Real-World Modeling**

Some data structures directly model real-world situations. The most important data structure of this type is the graph. You can use graphs to represent airline routes between cities or connections in an electric circuit or tasks in a project. We’ll cover graphs in Chapter 13, “Graphs,” and Chapter 14, “Weighted Graphs.” Other data structures, such as stacks and queues, may also be used in simulations. A queue, for example, can model customers waiting in line at a bank or cars waiting at a toll booth.

**Overview of Data Structures**

Another way to look at data structures is to focus on their strengths and weaknesses.

In this section we’ll provide an overview, in the form of a table, of the major data storage structures we’ll be discussing in this book. This is a bird’s-eye view of a land-scape that we’ll be covering later at ground level, so don’t be alarmed if the terms used are not familiar. Table 1.1 shows the advantages and disadvantages of the various data structures described in this book.

**TABLE 1.1**

Characteristics of Data Structures

**Data Structure**

**Advantages**

**Disadvantages**

Array

Quick insertion, very

Slow search,

fast access if index

slow deletion,

known.

fixed size.

Ordered array

Quicker search than

Slow insertion and

unsorted array.

deletion, fixed size.

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12

**CHAPTER 1**

Overview

**TABLE 1.1**

Continued

**Data Structure**

**Advantages**

**Disadvantages**

Stack

Provides last-in,

Slow access to

first-out access.

other items.

Queue

Provides first-in,

Slow access to

first-out access.

other items.

Linked list

Quick insertion,

Slow search.

quick deletion.

Binary tree

Quick search, insertion,

Deletion algorithm

deletion (if tree

is complex.

remains balanced).

Red-black tree

Quick search, insertion,

Complex.

deletion. Tree always

balanced.

2-3-4 tree

Quick search, insertion,

Complex.

deletion. Tree always

balanced. Similar trees

good for disk storage.

Hash table

Very fast access if

Slow deletion,

key known. Fast insertion.

access slow if key

not known, inefficient

memory usage.

Heap

Fast insertion, deletion,

Slow access to

access to largest item.

other items.

Graph

Models real-world

Some algorithms are

situations.

slow and complex.

The data structures shown in Table 1.1, except the arrays, can be thought of as Abstract Data Types, or ADTs. We’ll describe what this means in Chapter 5, “Linked Lists.”

**Overview of Algorithms**

Many of the algorithms we’ll discuss apply directly to specific data structures. For most data structures, you need to know how to

• Insert a new data item.

• Search for a specified item.

• Delete a specified item.

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Some Definitions

13

You may also need to know how to

Another important algorithm category is

The concept of

(The term

**Some Definitions**

Let’s look at a few of the terms that we’ll be using throughout this book.

**Database**

We’ll use the term

**Record**

**Field**

A record is usually divided into several

On an index card for an address book, a person’s name, address, or telephone number is an individual field.

More sophisticated database programs use records with more fields. Figure 1.1 shows such a record, where each line represents a distinct field.

In Java (and other object-oriented languages), records are usually represented by

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14

**CHAPTER 1**

Overview

Employee number:

Social security number:

Last name:

First name:

Street address:

City:

State:

Zip code:

Phone number:

Date of birth:

Date of first employment:

Salary:

**FIGURE 1.1**

A record with multiple fields.

**Key**

To search for a record within a database, you need to designate one of the record’s fields as a

You could search through the same file using the phone number field or the address field as the key. Any of the fields in Figure 1.1 could be used as a search key.

**Object-Oriented Programming**

This section is for those of you who haven’t been exposed to object-oriented programming. However, caveat emptor. We cannot, in a few pages, do justice to all the innovative new ideas associated with OOP. Our goal is merely to make it possible for you to understand the example programs in the text.

If, after reading this section and examining some of the example code in the following chapters, you still find the whole OOP business as alien as quantum physics, you may need a more thorough exposure to OOP. See the reading list in Appendix B,

“Further Reading,” for suggestions.

**Problems with Procedural Languages**

OOP was invented because procedural languages, such as C, Pascal, and early versions of BASIC, were found to be inadequate for large and complex programs.

Why was this?

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Object-Oriented Programming

15

There were two kinds of problems. One was the lack of correspondence between the program and the real world, and the other was the internal organization of the program.

**Poor Modeling of the Real World**

Conceptualizing a real-world problem using procedural languages is difficult.

Methods carry out a task, while data stores information, but most real-world objects do both of these things. The thermostat on your furnace, for example, carries out tasks (turning the furnace on and off) but also stores information (the current temperature and the desired temperature).

If you wrote a thermostat control program in a procedural language, you might end up with two methods, furnace_on() and furnace_off(), but also two global variables, currentTemp (supplied by a thermometer) and desiredTemp (set by the user).

However, these methods and variables wouldn’t form any sort of programming unit; there would be no unit in the program you could call thermostat. The only such concept would be in the programmer’s mind.

For large programs, which might contain hundreds of entities like thermostats, this procedural approach made things chaotic, error-prone, and sometimes impossible to implement at all. What was needed was a better match between things in the program and things in the outside world.

**Crude Organizational Units**

A more subtle, but related, problem had to do with a program’s internal organization. Procedural programs were organized by dividing the code into methods. One difficulty with this kind of method-based organization was that it focused on methods at the expense of data. There weren’t many options when it came to data.

To simplify slightly, data could be local to a particular method, or it could be global—accessible to all methods. There was no way (at least not a flexible way) to specify that some methods could access a variable and others couldn’t.

This inflexibility caused problems when several methods needed to access the same data. To be available to more than one method, such variables needed to be global, but global data could be accessed inadvertently by

**Objects in a Nutshell**

The idea of

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**CHAPTER 1**

Overview

**Objects**

Here’s the amazing breakthrough that is the key to OOP: An object contains

This new entity, the object, solves several problems simultaneously. Not only does an object in a program correspond more closely to an object in the real world, but it also solves the problem engendered by global data in the procedural model. The furnace_on() and furnace_off() methods can access currentTemp and desiredTemp.

These variables are hidden from methods that are not part of thermostat, however, so they are less likely to be accidentally changed by a rogue method.

**Classes**

You might think that the idea of an object would be enough for one programming revolution, but there’s more. Early on, it was realized that you might want to make several objects of the same type. Maybe you’re writing a furnace control program for an entire apartment building, for example, and you need several dozen thermostat objects in your program. It seems a shame to go to the trouble of specifying each one separately. Thus, the idea of classes was born.

A

class thermostat

{

private float currentTemp();

private float desiredTemp();

public void furnace_on()

{

// method body goes here

}

public void furnace_off()

{

// method body goes here

}

} // end class thermostat

The Java keyword class introduces the class specification, followed by the name you want to give the class; here it’s thermostat. Enclosed in curly brackets are the fields and methods that make up the class. We’ve left out the bodies of the methods; normally, each would have many lines of program code.

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Object-Oriented Programming

17

C programmers will recognize this syntax as similar to a structure, while C++

programmers will notice that it’s very much like a class in C++, except that there’s no semicolon at the end. (Why did we need the semicolon in C++ anyway?) **Creating Objects**

Specifying a class doesn’t create any objects of that class. (In the same way, specifying a structure in C doesn’t create any variables.) To actually create objects in Java, you must use the keyword new. At the same time an object is created, you need to store a reference to it in a variable of suitable type—that is, the same type as the class.

What’s a reference? We’ll discuss references in more detail later. In the meantime, think of a reference as a name for an object. (It’s actually the object’s address, but you don’t need to know that.)

Here’s how we would create two references to type thermostat, create two new thermostat objects, and store references to them in these variables: thermostat therm1, therm2; // create two references therm1 = new thermostat(); // create two objects and therm2 = new thermostat(); // store references to them Incidentally, creating an object is also called

**Accessing Object Methods**

After you specify a class and create some objects of that class, other parts of your program need to interact with these objects. How do they do that?

Typically, other parts of the program interact with an object’s methods, not with its data (fields). For example, to tell the therm2 object to turn on the furnace, we would say

therm2.furnace_on();

The dot operator (.) associates an object with one of its methods (or occasionally with one of its fields).

At this point we’ve covered (rather telegraphically) several of the most important features of OOP. To summarize:

• Objects contain both methods and fields (data).

• A class is a specification for any number of objects.

• To create an object, you use the keyword new in conjunction with the class name.

• To invoke a method for a particular object, you use the dot operator.

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**CHAPTER 1**

Overview

These concepts are deep and far reaching. It’s almost impossible to assimilate them the first time you see them, so don’t worry if you feel a bit confused. As you see more classes and what they do, the mist should start to clear.

**A Runnable Object-Oriented Program**

Let’s look at an object-oriented program that runs and generates actual output. It features a class called BankAccount that models a checking account at a bank. The program creates an account with an opening balance, displays the balance, makes a deposit and a withdrawal, and then displays the new balance. Listing 1.1 shows bank.java.

**LISTING 1.1**

The bank.java Program

// bank.java

// demonstrates basic OOP syntax

// to run this program: C>java BankApp

////////////////////////////////////////////////////////////////

class BankAccount

{

private double balance; // account balance public BankAccount(double openingBalance) // constructor

{

balance = openingBalance;

}

public void deposit(double amount) // makes deposit

{

balance = balance + amount;

}

public void withdraw(double amount) // makes withdrawal

{

balance = balance - amount;

}

public void display() // displays balance

{

System.out.println(“balance=” + balance);

}

} // end class BankAccount

////////////////////////////////////////////////////////////////

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Object-Oriented Programming

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**LISTING 1.1**

Continued

class BankApp

{

public static void main(String[] args)

{

BankAccount ba1 = new BankAccount(100.00); // create acct System.out.print(“Before transactions, “);

ba1.display(); // display balance ba1.deposit(74.35); // make deposit ba1.withdraw(20.00); // make withdrawal System.out.print(“After transactions, “);

ba1.display(); // display balance

} // end main()

} // end class BankApp

Here’s the output from this program:

Before transactions, balance=100

After transactions, balance=154.35

There are two classes in bank.java. The first one, BankAccount, contains the fields and methods for our bank account. We’ll examine it in detail in a moment. The second class, BankApp, plays a special role.

**The **BankApp **Class**

To execute the program in Listing 1.1 from an MS-DOS prompt, you type java BankApp following the C: prompt:

C:\> **java BankApp**

This command tells the java interpreter to look in the BankApp class for the method called main(). Every Java application must have a main() method; execution of the program starts at the beginning of main(), as you can see in Listing 1.1. (You don’t need to worry yet about the String[] args argument in main().) The main() method creates an object of class BankAccount, initialized to a value of 100.00, which is the opening balance, with this statement: BankAccount ba1 = new BankAccount(100.00); // create acct

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**CHAPTER 1**

Overview

The System.out.print() method displays the string used as its argument, Before transactions:, and the account displays its balance with this statement: ba1.display();

The program then makes a deposit to, and a withdrawal from, the account: ba1.deposit(74.35);

ba1.withdraw(20.00);

Finally, the program displays the new account balance and terminates.

**The **BankAccount **Class**

The only data field in the BankAccount class is the amount of money in the account, called balance. There are three methods. The deposit() method adds an amount to the balance, withdrawal() subtracts an amount, and display() displays the balance.

**Constructors**

The BankAccount class also features a

A constructor allows a new object to be initialized in a convenient way. Without the constructor in this program, you would have needed an additional call to deposit() to put the opening balance in the account.

**Public and Private**

Notice the keywords public and private in the BankAccount class. These keywords are

All the methods in BankAccount have the access modifier public, however, so they can be accessed by methods in other classes. That’s why statements in main() can call deposit(), withdrawal(), and display().

Data fields in a class are typically made private and methods are made public. This protects the data; it can’t be accidentally modified by methods of other classes. Any outside entity that needs to access data in a class must do so using a method of the same class. Data is like a queen bee, kept hidden in the middle of the hive, fed and cared for by worker-bee methods.

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Software Engineering

21

**Inheritance and Polymorphism**

We’ll briefly mention two other key features of object-oriented programming: inheritance and polymorphism.

In Java, inheritance is also called

Inheritance enables you to easily add features to an existing class and is an important aid in the design of programs with many related classes. Inheritance thus makes it easy to reuse classes for a slightly different purpose, a key benefit of OOP.

In practice, polymorphism usually involves a method call that actually executes different methods for objects of different classes.

For example, a call to display() for a secretary object would invoke a display method in the secretary class, while the exact same call for a manager object would invoke a different display method in the manager class. Polymorphism simplifies and clarifies program design and coding.

For those not familiar with them, inheritance and polymorphism involve significant additional complexity. To keep the focus on data structures and algorithms, we have avoided these features in our example programs. Inheritance and polymorphism are important and powerful aspects of OOP but are not necessary for the explanation of data structures and algorithms.

**Software Engineering**

In recent years, it has become fashionable to begin a book on data structures and algorithms with a chapter on software engineering. We don’t follow that approach, but let’s briefly examine software engineering and see how it fits into the topics we discuss in this book.

Software engineering is the study of ways to create large and complex computer programs, involving many programmers. It focuses on the overall design of the programs and on the creation of that design from the needs of the end users.

Software engineering is concerned with the life cycle of a software project, which includes specification, design, verification, coding, testing, production, and maintenance.

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**CHAPTER 1**

Overview

It’s not clear that mixing software engineering on one hand and data structures and algorithms on the other actually helps the student understand either topic. Software engineering is rather abstract and is difficult to grasp until you’ve been involved yourself in a large project. The use of data structures and algorithms, on the other hand, is a nuts-and-bolts discipline concerned with the details of coding and data storage.

Accordingly, we focus on the essentials of data structures and algorithms. How do they really work? What structure or algorithm is best in a particular situation? What do they look like translated into Java code? As we noted, our intent is to make the material as easy to understand as possible. For further reading, we mention some books on software engineering in Appendix B.

**Java for C++ Programmers**

If you’re a C++ programmer who has not yet encountered Java, you might want to read this section. We’ll mention several ways that Java differs from C++.

This section is not intended to be a primer on Java. We don’t even cover all the differences between C++ and Java. We’re interested in only a few Java features that might make it hard for C++ programmers to figure out what’s going on in the example programs.

**No Pointers**

The biggest difference between C++ and Java is that Java doesn’t use pointers. To a C++ programmer, not using pointers may at first seem quite amazing. How can you get along without pointers?

Throughout this book we’ll use pointer-free code to build complex data structures.

You’ll see that this approach is not only possible, but actually easier than using C++

pointers.

Actually, Java only does away with

**References**

Java treats primitive data types (such as int, float, and double) differently than objects. Look at these two statements:

int intVar; // an int variable called intVar

BankAccount bc1; // reference to a BankAccount object

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Java for C++ Programmers

23

In the first statement, a memory location called intVar actually holds a numerical value such as 127 (assuming such a value has been placed there). However, the memory location bc1 does not hold the data of a BankAccount object. Instead, it contains the

Actually, bc1 won’t hold a reference if it has not been assigned an object at some prior point in the program. Before being assigned an object, it holds a reference to a special object called null. In the same way, intVar won’t hold a numerical value if it’s never been assigned one. The compiler will complain if you try to use a variable that has never been assigned a value.

In C++, the statement

BankAccount bc1;

actually creates an object; it sets aside enough memory to hold all the object’s data.

In Java, all this statement creates is a place to put an object’s memory address. You can think of a reference as a pointer with the syntax of an ordinary variable. (C++

has reference variables, but they must be explicitly specified with the & symbol.) **Assignment**

It follows that the assignment operator (=) operates differently with Java objects than with C++ objects. In C++, the statement

bc2 = bc1;

copies all the data from an object called bc1 into a different object called bc2.

Following this statement, there are two objects with the same data. In Java, on the other hand, this same assignment statement copies the memory address that bc1

refers to into bc2. Both bc1 and bc2 now refer to exactly the same object; they are references to it.

This can get you into trouble if you’re not clear what the assignment operator does.

Following the assignment statement shown above, the statement bc1.withdraw(21.00);

and the statement

bc2.withdraw(21.00);

both withdraw $21 from

Suppose you actually want to copy data from one object to another. In this case you must make sure you have two separate objects to begin with and then copy each field separately. The equal sign won’t do the job.

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**CHAPTER 1**

Overview

**The **new **Operator**

Any object in Java must be created using new. However, in Java, new returns a reference, not a pointer as in C++. Thus, pointers aren’t necessary to use new. Here’s one way to create an object:

BankAccount ba1;

ba1 = new BankAccount();

Eliminating pointers makes for a more secure system. As a programmer, you can’t find out the actual address of ba1, so you can’t accidentally corrupt it. However, you probably don’t need to know it, unless you’re planning something wicked.

How do you release memory that you’ve acquired from the system with new and no longer need? In C++, you use delete. In Java, you don’t need to worry about releas-ing memory. Java periodically looks through each block of memory that was obtained with new to see if valid references to it still exist. If there are no such references, the block is returned to the free memory store. This process is called

In C++ almost every programmer at one time or another forgets to delete memory blocks, causing “memory leaks” that consume system resources, leading to bad performance and even crashing the system. Memory leaks can’t happen in Java (or at least hardly ever).

**Arguments**

In C++, pointers are often used to pass objects to functions to avoid the overhead of copying a large object. In Java, objects are always passed as references. This approach also avoids copying the object:

void method1()

{

BankAccount ba1 = new BankAccount(350.00);

method2(ba1);

}

void method2(BankAccount acct)

{

}

In this code, the references ba1 and acct both refer to the same object. In C++ acct would be a separate object, copied from ba1.

Primitive data types, on the other hand, are always passed by value. That is, a new variable is created in the method and the value of the argument is copied into it.

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Java for C++ Programmers

25

**Equality and Identity**

In Java, if you’re talking about primitive types, the equality operator (==) will tell you whether two variables have the same value:

int intVar1 = 27;

int intVar2 = intVar1;

if(intVar1 == intVar2)

System.out.println(“They’re equal”);

This is the same as the syntax in C and C++, but in Java, because relational operators use references, they work differently with objects. The equality operator, when applied to objects, tells you whether two references are identical—that is, whether they refer to the same object:

carPart cp1 = new carPart(“fender”);

carPart cp2 = cp1;

if(cp1 == cp2)

System.out.println(“They’re Identical”);

In C++ this operator would tell you if two objects contained the same data. If you want to see whether two objects contain the same data in Java, you must use the equals() method of the Object class:

carPart cp1 = new carPart(“fender”);

carPart cp2 = cp1;

if( cp1.equals(cp2) )

System.out.println(“They’re equal”);

This technique works because all objects in Java are implicitly derived from the Object class.

**Overloaded Operators**

This point is easy: There are no overloaded operators in Java. In C++, you can redefine +, *, =, and most other operators so that they behave differently for objects of a particular class. No such redefinition is possible in Java. Use a named method instead, such as add() or whatever.

**Primitive Variable Types**

The primitive or built-in variable types in Java are shown in Table 1.2.

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**CHAPTER 1**

Overview

**TABLE 1.2**

Primitive Data Types

**Name**

**Size in Bits**

**Range of Values**

boolean

1

true or false

byte

8

–128 to +127

char

16

‘\u0000’ to ‘\uFFFF’

short

16

–32,768 to +32,767

int

32

–2,147,483,648 to +2,147,483,647

long

64

–9,223,372,036,854,775,808 to

+9,223,372,036,854,775,807

float

32

Approximately 10–38 to 10+38; 7 significant digits

double

64

Approximately 10–308 to 10+308; 15 significant digits Unlike C and C++, which use integers for true/false values, boolean is a distinct type in Java.

Type char is unsigned, and uses two bytes to accommodate the Unicode character representation scheme, which can handle international characters.

The int type varies in size in C and C++, depending on the specific computer platform; in Java an int is always 32 bits.

Literals of type float use the suffix F (for example, 3.14159F); literals of type double need no suffix. Literals of type long use suffix L (as in 45L); literals of the other integer types need no suffix.

Java is more strongly typed than C and C++; many conversions that were automatic in those languages require an explicit cast in Java.

All types not shown in Table 1.2, such as String, are classes.

**Input/Output**

There have been changes to input/output as Java has evolved. For the console-mode applications we’ll be using as example programs in this book, some clunky-looking but effective constructions are available for input and output. They’re quite different from the workhorse cout and cin approaches in C++ and printf() and scanf() in C.

Older versions of the Java Software Development Kit (SDK) required the line import java.io.*;

at the beginning of the source file for all input/output routines. Now this line is needed only for input.

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Java for C++ Programmers

27

**Output**

You can send any primitive type (numbers and characters), and String objects as well, to the display with these statements:

System.out.print(var); // displays var, no linefeed System.out.println(var); // displays var, then starts new line The print() method leaves the cursor on the same line; println() moves it to the beginning of the next line.

In older versions of the SDK, a System.out.print() statement did not actually write anything to the screen. It had to be followed by a System.out.println()or System.out.flush() statement to display the entire buffer. Now it displays immediately.

You can use several variables, separated by plus signs, in the argument. Suppose in this statement the value of ans is 33:

System.out.println(“The answer is “ + ans);

Then the output will be

The answer is 33

**Inputting a String**

Input is considerably more involved than output. In general, you want to read any input as a String object. If you’re actually inputting something else, say a character or number, you then convert the String object to the desired type.

As we noted, any program that uses input must include the statement import java.io.*;

at the beginning of the program. Without this statement, the compiler will not recognize such entities as IOException and InputStreamReader.

String input is fairly baroque. Here’s a method that returns a string entered by the user:

public static String getString() throws IOException

{

InputStreamReader isr = new InputStreamReader(System.in); BufferedReader br = new BufferedReader(isr);

String s = br.readLine();

return s;

}

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**CHAPTER 1**

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This method returns a String object, which is composed of characters typed on the keyboard and terminated with the Enter key. The details of the InputStreamReader and BufferedReader classes need not concern us here.

Besides importing java.io.*, you’ll need to add throws IOException to all input methods, as shown in the preceding code. In fact, you’ll need to add throws IOException to any method, such as main(), that calls any of the input methods.

**Inputting a Character**

Suppose you want your program’s user to enter a character. (By

{

String s = getString();

return s.charAt(0);

}

The charAt() method of the String class returns a character at the specified position in the String object; here we get the first character, which is number 0. This approach prevents extraneous characters being left in the input buffer. Such characters can cause problems with subsequent input.

**Inputting Integers**

To read numbers, you make a String object as shown before and convert it to the type you want using a conversion method. Here’s a method, getInt(), that converts input into type int and returns it:

public int getInt() throws IOException

{

String s = getString();

return Integer.parseInt(s);

}

The parseInt() method of class Integer converts the string to type int. A similar routine, parseLong(), can be used to convert type long.

In older versions of the SDK, you needed to use the line import java.lang.Integer;

at the beginning of any program that used parseInt(), but this convention is no longer necessary.

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Java Library Data Structures

29

For simplicity, we don’t show any error-checking in the input routines in the example programs. The user must type appropriate input, or an exception will occur.

With the code shown here the exception will cause the program to terminate. In a serious program you should analyze the input string before attempting to convert it and should also catch any exceptions and process them appropriately.

**Inputting Floating-Point Numbers**

Types float and double can be handled in somewhat the same way as integers, but the conversion process is more complex. Here’s how you read a number of type double:

public int getDouble() throws IOException

{

String s = getString();

Double aDub = Double.valueOf(s);

return aDub.doubleValue();

}

The String is first converted to an object of type Double (uppercase

“wrapper” class for type double. A method of Double called doubleValue() then converts the object to type double.

For type float, there’s an equivalent Float class, which has equivalent valueOf() and floatValue() methods.

**Java Library Data Structures**

The java.util package contains data structures, such as Vector (an extensible array), Stack, Dictionary, and Hashtable. In this book we’ll usually ignore these built-in classes. We’re interested in teaching fundamentals, not the details of a particular implementation. However, occasionally we’ll find some of these structures useful.

You must use the line

import java.util.*;

before you can use objects of these classes.

Although we don’t focus on them, such class libraries, whether those that come with Java or others available from third-party developers, can offer a rich source of versatile, debugged storage classes. This book should equip you with the knowledge to know what sort of data structure you need and the fundamentals of how it works.

Then you can decide whether you should write your own classes or use someone else’s.

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**CHAPTER 1**

Overview

**Summary**

• A data structure is the organization of data in a computer’s memory or in a disk file.

• The correct choice of data structure allows major improvements in program efficiency.

• Examples of data structures are arrays, stacks, and linked lists.

• An algorithm is a procedure for carrying out a particular task.

• In Java, an algorithm is usually implemented by a class method.

• Many of the data structures and algorithms described in this book are most often used to build databases.

• Some data structures are used as programmer’s tools: They help execute an algorithm.

• Other data structures model real-world situations, such as telephone lines running between cities.

• A database is a unit of data storage composed of many similar records.

• A record often represents a real-world object, such as an employee or a car part.

• A record is divided into fields. Each field stores one characteristic of the object described by the record.

• A key is a field in a record that’s used to carry out some operation on the data.

For example, personnel records might be sorted by a LastName field.

• A database can be searched for all records whose key field has a certain value.

This value is called a search key.

**Questions**

These questions are intended as a self-test for readers. Answers to the questions may be found in Appendix C.

**1. **In many data structures you can ________ a single record, _________ it, and _______ it.

**2. **Rearranging the contents of a data structure into a certain order is called _________ .

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Questions

31

**3. **In a database, a field is

**a. **a specific data item.

**b. **a specific object.

**c. **part of a record.

**d. **part of an algorithm.

**4. **The field used when searching for a particular record is the ______________ .

**5. **In object-oriented programming, an object **a. **is a class.

**b. **may contain data and methods.

**c. **is a program.

**d. **may contain classes.

**6. **A class

**a. **is a blueprint for many objects.

**b. **represents a specific real-world object.

**c. **will hold specific values in its fields.

**d. **specifies the type of a method.

**7. **In Java, a class specification

**a. **creates objects.

**b. **requires the keyword new.

**c. **creates references.

**d. **none of the above.

**8. **When an object wants to do something, it uses a ________ .

**9. **In Java, accessing an object’s methods requires the _____ operator.

**10. **In Java, boolean and byte are _____________ .

(There are no experiments or programming projects for Chapter 1.)

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**2**

**IN THIS CHAPTER**

• The Basics of Arrays in Java

Arrays

• Dividing a Program into

Classes

• Class Interfaces

The array is the most commonly used data storage struc-

• Java Code for an Ordered

ture; it’s built into most programming languages. Because arrays are so well known, they offer a convenient jumping-Array

off place for introducing data structures and for seeing

• Logarithms

how object-oriented programming and data structures relate to one another. In this chapter we’ll introduce arrays

• Storing Objects

in Java and demonstrate a home-made array class.

• Big O Notation

We’ll also examine a special kind of array, the ordered

• Why Not Use Arrays for

array, in which the data is stored in ascending (or descend-Everything?

ing) key order. This arrangement makes possible a fast way of searching for a data item: the binary search.

We’ll start the chapter with a Java Workshop applet that shows insertion, searching, and deletion in an array. Then we’ll show some sample Java code that carries out these same operations.

Later we’ll examine ordered arrays, again starting with a Workshop applet. This applet will demonstrate a binary search. At the end of the chapter we’ll talk about Big O

notation, the most widely used measure of algorithm efficiency.

**The Array Workshop Applet**

Suppose you’re coaching kids-league baseball, and you want to keep track of which players are present at the practice field. What you need is an attendance-monitoring program for your laptop—a program that maintains a database of the players who have shown up for practice. You can use a simple data structure to hold this data. There are several actions you would like to be able to perform:

• Insert a player into the data structure when the

player arrives at the field.

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• Check to see whether a particular player is present, by searching for the player’s number in the structure.

• Delete a player from the data structure when that player goes home.

These three operations—insertion, searching, and deletion—will be the fundamental ones in most of the data storage structures we’ll study in this book.

We’ll often begin the discussion of a particular data structure by demonstrating it with a Workshop applet. This approach will give you a feeling for what the structure and its algorithms do, before we launch into a detailed explanation and demonstrate sample code. The Workshop applet called Array shows how an array can be used to implement insertion, searching, and deletion.

Now start up the Array Workshop applet, as described in Appendix A, “Running the Workshop Applets and Example Programs,” with

C:\> **appletviewer Array.html**

Figure 2.1 shows the resulting array with 20 elements, 10 of which have data items in them. You can think of these items as representing your baseball players. Imagine that each player has been issued a team shirt with the player’s number on the back.

To make things visually interesting, the shirts come in a variety of colors. You can see each player’s number and shirt color in the array.

**FIGURE 2.1**

The Array Workshop applet.

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This applet demonstrates the three fundamental procedures mentioned earlier:

• The Ins button inserts a new data item.

• The Find button searches for specified data item.

• The Del button deletes a specified data item.

Using the New button, you can create a new array of a size you specify. You can fill this array with as many data items as you want using the Fill button. Fill creates a set of items and randomly assigns them numbers and colors. The numbers are in the range 0 to 999. You can’t create an array of more than 60 cells, and you can’t, of course, fill more data items than there are array cells.

Also, when you create a new array, you’ll need to decide whether duplicate items will be allowed; we’ll return to this question in a moment. The default value is no duplicates, so the No Dups radio button is initially selected to indicate this setting.

**Insertion**

Start with the default arrangement of 20 cells and 10 data items, and the No Dups button selected. You insert a baseball player’s number into the array when the player arrives at the practice field, having been dropped off by a parent. To insert a new item, press the Ins button once. You’ll be prompted to enter the value of the item: Enter key of item to insert

Type a number, say 678, into the text field in the upper-right corner of the applet.

(Yes, it is hard to get three digits on the back of a kid’s shirt.) Press Ins again and the applet will confirm your choice:

Will insert item with key 678

A final press of the button will cause a data item, consisting of this value and a random color, to appear in the first empty cell in the array. The prompt will say something like

Inserted item with key 678 at index 10

Each button press in a Workshop applet corresponds to a step that an algorithm carries out. The more steps required, the longer the algorithm takes. In the Array Workshop applet the insertion process is very fast, requiring only a single step. This is true because a new item is always inserted in the first vacant cell in the array, and the algorithm knows this location because it knows how many items are already in the array. The new item is simply inserted in the next available space. Searching and deletion, however, are not so fast.

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In no-duplicates mode you’re on your honor not to insert an item with the same key as an existing item. If you do, the applet displays an error message, but it won’t prevent the insertion. The assumption is that you won’t make this mistake.

**Searching**

To begin a search, click the Find button. You’ll be prompted for the key number of the person you’re looking for. Pick a number that appears on an item somewhere in the middle of the array. Type in the number and repeatedly press the Find button. At each button press, one step in the algorithm is carried out. You’ll see the red arrow start at cell 0 and move methodically down the cells, examining a new one each time you press the button. The index number in the message Checking next cell, index = 2

will change as you go along. When you reach the specified item, you’ll see the message

Have found item with key 505

or whatever key value you typed in. Assuming duplicates are not allowed, the search will terminate as soon as an item with the specified key value is found.

If you have selected a key number that is not in the array, the applet will examine every occupied cell in the array before telling you that it can’t find that item.

Notice that (again assuming duplicates are not allowed) the search algorithm must look through an average of half the data items to find a specified item. Items close to the beginning of the array will be found sooner, and those toward the end will be found later. If N is the number of items, the average number of steps needed to find an item is N/2. In the worst-case scenario, the specified item is in the last occupied cell, and N steps will be required to find it.

As we noted, the time an algorithm takes to execute is proportional to the number of steps, so searching takes much longer on the average (N/2 steps) than insertion (one step).

**Deletion**

To delete an item, you must first find it. After you type in the number of the item to be deleted, repeated button presses will cause the arrow to move, step by step, down the array until the item is located. The next button press deletes the item, and the cell becomes empty. (Strictly speaking, this step isn’t necessary because we’re going to copy over this cell anyway, but deleting the item makes it clearer what’s happening.)

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37

Implicit in the deletion algorithm is the assumption that

Therefore, after locating the specified item and deleting it, the applet must shift the contents of each subsequent cell down one space to fill in the hole. Figure 2.2 shows an example.

Item to be

deleted

0

1

2

3

4

5

6

7

8

9

84

61

15

73

26

38

11

49

53

32

❶

❷

❸

❹

0

1

2

3

4

5

6

7

8

84

61

15

73

26

11

49

53

32

Contents

shifted

down

**FIGURE 2.2**

Deleting an item.

If the item in cell 5 (38, in Figure 2.2) is deleted, the item in 6 shifts into 5, the item in 7 shifts into 6, and so on to the last occupied cell. During the deletion process, when the item is located, the applet shifts down the contents of the higher-indexed cells as you continue to press the Del button.

A deletion requires (assuming no duplicates are allowed) searching through an average of N/2 elements and then moving the remaining elements (an average of N/2 moves) to fill up the resulting hole. This is N steps in all.

**The Duplicates Issue**

When you design a data storage structure, you need to decide whether items with duplicate keys will be allowed. If you’re working with a personnel file and the key is an employee number, duplicates don’t make much sense; there’s no point in assign-ing the same number to two employees. On the other hand, if the key value is last names, then there’s a distinct possibility several employees will have the same key value, so duplicates should be allowed.

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Of course, for the baseball players, duplicate numbers should not be allowed.

Keeping track of the players would be hard if more than one wore the same number.

The Array Workshop applet lets you select either option. When you use New to create a new array, you’re prompted to specify both its size and whether duplicates are permitted. Use the radio buttons Dups OK or No Dups to make this selection.

If you’re writing a data storage program in which duplicates are not allowed, you may need to guard against human error during an insertion by checking all the data items in the array to ensure that none of them already has the same key value as the item being inserted. This check is inefficient, however, and increases the number of steps required for an insertion from one to N. For this reason, our applet does not perform this check.

**Searching with Duplicates**

Allowing duplicates complicates the search algorithm, as we noted. Even if it finds a match, it must continue looking for possible additional matches until the last occupied cell. At least, this is one approach; you could also stop after the first match.

How you proceed depends on whether the question is “Find me everyone with blue eyes” or “Find me someone with blue eyes.”

When the Dups OK button is selected, the applet takes the first approach, finding all items matching the search key. This approach always requires N steps because the algorithm must go all the way to the last occupied cell.

**Insertion with Duplicates**

Insertion is the same with duplicates allowed as when they’re not: A single step inserts the new item. But remember, if duplicates are not allowed, and there’s a possibility the user will attempt to input the same key twice, you may need to check every existing item before doing an insertion.

**Deletion with Duplicates**

Deletion may be more complicated when duplicates are allowed, depending on exactly how “deletion” is defined. If it means to delete only the first item with a specified value, then, on the average, only N/2 comparisons and N/2 moves are necessary. This is the same as when no duplicates are allowed.

If, however, deletion means to delete

cells and (probably) moving more than N/2 cells. The average depends on how the duplicates are distributed throughout the array.

The applet assumes this second meaning and deletes multiple items with the same key. This is complicated because each time an item is deleted, subsequent items must be shifted farther. For example, if three items are deleted, then items beyond the last

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39

deletion will need to be shifted three spaces. To see how this operation works, set the applet to Dups OK and insert three or four items with the same key. Then try deleting them.

Table 2.1 shows the average number of comparisons and moves for the three operations, first where no duplicates are allowed and then where they are allowed. N is the number of items in the array. Inserting a new item counts as one move.

**TABLE 2.1**

Duplicates OK Versus No Duplicates

**No Duplicates**

**Duplicates OK**

Search

N/2 comparisons

N comparisons

Insertion

No comparisons, one move

No comparisons, one move

Deletion

N/2 comparisons, N/2 moves

N comparisons, more than N/2 moves

You can explore these possibilities with the Array Workshop applet.

The difference between N and N/2 is not usually considered very significant, except when you’re fine-tuning a program. Of more importance, as we’ll discuss toward the end of this chapter, is whether an operation takes one step, N steps, log(N) steps, or N2 steps.

**Not Too Swift**

One of the significant things to notice when you’re using the Array applet is the slow and methodical nature of the algorithms. With the exception of insertion, the algorithms involve stepping through some or all of the cells in the array. Different data structures offer much faster (but more complex) algorithms. We’ll see one, the binary search on an ordered array, later in this chapter, and others throughout this book.

**The Basics of Arrays in Java**

The preceding section showed graphically the primary algorithms used for arrays.

Now we’ll see how to write programs to carry out these algorithms, but we first want to cover a few of the fundamentals of arrays in Java.

If you’re a Java expert, you can skip ahead to the next section, but even C and C++

programmers should stick around. Arrays in Java use syntax similar to that in C and C++ (and not that different from other languages), but there are nevertheless some unique aspects to the Java approach.

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**Creating an Array**

As we noted in Chapter 1, “Overview,” there are two kinds of data in Java: primitive types (such as int and double) and objects. In many programming languages (even object-oriented ones such as C++), arrays are primitive types, but in Java they’re treated as objects. Accordingly, you must use the new operator to create an array: int[] intArray; // defines a reference to an array intArray = new int[100]; // creates the array, and

// sets intArray to refer to it

Or you can use the equivalent single-statement approach: int[] intArray = new int[100];

The [] operator is the sign to the compiler we’re naming an array object and not an ordinary variable. You can also use an alternative syntax for this operator, placing it after the name instead of the type:

int intArray[] = new int[100]; // alternative syntax However, placing the [] after the int makes it clear that the [] is part of the type, not the name.

Because an array is an object, its name—intArray in the preceding code—is a reference to an array; it’s not the array itself. The array is stored at an address elsewhere in memory, and intArray holds only this address.

Arrays have a length field, which you can use to find the size (the number of elements) of an array:

int arrayLength = intArray.length; // find array size As in most programming languages, you can’t change the size of an array after it’s been created.

**Accessing Array Elements**

Array elements are accessed using an index number in square brackets. This is similar to how other languages work:

temp = intArray[3]; // get contents of fourth element of array intArray[7] = 66; // insert 66 into the eighth cell Remember that in Java, as in C and C++, the first element is numbered 0, so that the indices in an array of 10 elements run from 0 to 9.

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If you use an index that’s less than 0 or greater than the size of the array less 1, you’ll get the Array Index Out of Bounds runtime error.

**Initialization**

Unless you specify otherwise, an array of integers is automatically initialized to 0

when it’s created. Unlike C++, this is true even of arrays defined within a method (function). Say you create an array of objects like this: autoData[] carArray = new autoData[4000];

Until the array elements are given explicit values, they contain the special null object. If you attempt to access an array element that contains null, you’ll get the runtime error Null Pointer Assignment. The moral is to make sure you assign something to an element before attempting to access it.

You can initialize an array of a primitive type to something besides 0 using this syntax:

int[] intArray = { 0, 3, 6, 9, 12, 15, 18, 21, 24, 27 }; Perhaps surprisingly, this single statement takes the place of both the reference declaration and the use of new to create the array. The numbers within the curly brackets are called the

**An Array Example**

Let’s look at some example programs that show how an array can be used. We’ll start with an old-fashioned procedural version and then show the equivalent object-oriented approach. Listing 2.1 shows the old-fashioned version, called array.java.

**LISTING 2.1**

The array.java Program

// array.java

// demonstrates Java arrays

// to run this program: C>java arrayApp

////////////////////////////////////////////////////////////////

class ArrayApp

{

public static void main(String[] args)

{

long[] arr; // reference to array

arr = new long[100]; // make array

int nElems = 0; // number of items

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**LISTING 2.1**

Continued

int j; // loop counter

long searchKey; // key of item to search for

//--------------------------------------------------------------

arr[0] = 77; // insert 10 items

arr[1] = 99;

arr[2] = 44;

arr[3] = 55;

arr[4] = 22;

arr[5] = 88;

arr[6] = 11;

arr[7] = 00;

arr[8] = 66;

arr[9] = 33;

nElems = 10; // now 10 items in array

//--------------------------------------------------------------

for(j=0; j<nElems; j++) // display items

System.out.print(arr[j] + “ “);

System.out.println(“”);

//--------------------------------------------------------------

searchKey = 66; // find item with key 66

for(j=0; j<nElems; j++) // for each element, if(arr[j] == searchKey) // found item?

break; // yes, exit before end

if(j == nElems) // at the end?

System.out.println(“Can’t find “ + searchKey); // yes else

System.out.println(“Found “ + searchKey); // no

//--------------------------------------------------------------

searchKey = 55; // delete item with key 55

for(j=0; j<nElems; j++) // look for it if(arr[j] == searchKey)

break;

for(int k=j; k<nElems-1; k++) // move higher ones down arr[k] = arr[k+1];

nElems--; // decrement size

//--------------------------------------------------------------

for(j=0; j<nElems; j++) // display items

System.out.print( arr[j] + “ “);

System.out.println(“”);

} // end main()

} // end class ArrayApp

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In this program, we create an array called arr, place 10 data items (kids’ numbers) in it, search for the item with value 66 (the shortstop, Louisa), display all the items, remove the item with value 55 (Freddy, who had a dentist appointment), and then display the remaining 9 items. The output of the program looks like this: 77 99 44 55 22 88 11 0 66 33

Found 66

77 99 44 22 88 11 0 66 33

The data we’re storing in this array is type long. We use long to make it clearer that this is data; type int is used for index values. We’ve chosen a primitive type to simplify the coding. Generally, the items stored in a data structure consist of several fields, so they are represented by objects rather than primitive types. We’ll see such an example toward the end of this chapter.

**Insertion**

Inserting an item into the array is easy; we use the normal array syntax: arr[0] = 77;

We also keep track of how many items we’ve inserted into the array with the nElems variable.

**Searching**

The searchKey variable holds the value we’re looking for. To search for an item, we step through the array, comparing searchKey with each element. If the loop variable j reaches the last occupied cell with no match being found, the value isn’t in the array. Appropriate messages are displayed: Found 66 or Can’t find 27.

**Deletion**

Deletion begins with a search for the specified item. For simplicity, we assume (perhaps rashly) that the item is present. When we find it, we move all the items with higher index values down one element to fill in the “hole” left by the deleted element, and we decrement nElems. In a real program, we would also take appropriate action if the item to be deleted could not be found.

**Display**

Displaying all the elements is straightforward: We step through the array, accessing each one with arr[j] and displaying it.

**Program Organization**

The organization of array.java leaves something to be desired. The program has only one class, ArrayApp, and this class has only one method, main(). array.java is essentially an old-fashioned procedural program. Let’s see if we can make it easier to understand (among other benefits) by making it more object oriented.

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We’re going to provide a gradual introduction to an object-oriented approach, using two steps. In the first, we’ll separate the data storage structure (the array) from the rest of the program. The remaining part of the program will become a

**Dividing a Program into Classes**

The array.java program in Listing 2.1 essentially consists of one big method. We can reap many benefits by dividing the program into classes. What classes? The data storage structure itself is one candidate, and the part of the program that uses this data structure is another. By dividing the program into these two classes, we can clarify the functionality of the program, making it easier to design and understand (and in real programs to modify and maintain).

In array.java we used an array as a data storage structure, but we treated it simply as a language element. Now we’ll encapsulate the array in a class, called LowArray.

We’ll also provide class methods by which objects of other classes (the LowArrayApp class in this case) can access the array. These methods allow communication between LowArray and LowArrayApp.

Our first design of the LowArray class won’t be entirely successful, but it will demonstrate the need for a better approach. The lowArray.java program in Listing 2.2

shows how it looks.

**LISTING 2.2**

The lowArray.java Program

// lowArray.java

// demonstrates array class with low-level interface

// to run this program: C>java LowArrayApp

////////////////////////////////////////////////////////////////

class LowArray

{

private long[] a; // ref to array a

//--------------------------------------------------------------

public LowArray(int size) // constructor

{ a = new long[size]; } // create array

//--------------------------------------------------------------

public void setElem(int index, long value) // set value

{ a[index] = value; }

//--------------------------------------------------------------

public long getElem(int index) // get value

{ return a[index]; }

//--------------------------------------------------------------

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Dividing a Program into Classes

45

**LISTING 2.2**

Continued

} // end class LowArray

////////////////////////////////////////////////////////////////

class LowArrayApp

{

public static void main(String[] args)

{

LowArray arr; // reference

arr = new LowArray(100); // create LowArray object int nElems = 0; // number of items in array int j; // loop variable

arr.setElem(0, 77); // insert 10 items

arr.setElem(1, 99);

arr.setElem(2, 44);

arr.setElem(3, 55);

arr.setElem(4, 22);

arr.setElem(5, 88);

arr.setElem(6, 11);

arr.setElem(7, 00);

arr.setElem(8, 66);

arr.setElem(9, 33);

nElems = 10; // now 10 items in array for(j=0; j<nElems; j++) // display items

System.out.print(arr.getElem(j) + “ “);

System.out.println(“”);

int searchKey = 26; // search for data item for(j=0; j<nElems; j++) // for each element, if(arr.getElem(j) == searchKey) // found item?

break;

if(j == nElems) // no

System.out.println(“Can’t find “ + searchKey);

else // yes

System.out.println(“Found “ + searchKey);

// delete value 55

for(j=0; j<nElems; j++) // look for it if(arr.getElem(j) == 55)

break;

for(int k=j; k<nElems; k++) // higher ones down

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**LISTING 2.2**

Continued

arr.setElem(k, arr.getElem(k+1) );

nElems--; // decrement size for(j=0; j<nElems; j++) // display items

System.out.print( arr.getElem(j) + “ “);

System.out.println(“”);

} // end main()

} // end class LowArrayApp

////////////////////////////////////////////////////////////////

The output from the lowArray.java program is similar to that from array.java, except that we try to find a non-existent key value (26) before deleting the item with the key value 55:

77 99 44 55 22 88 11 0 66 33

Can’t find 26

77 99 44 22 88 11 0 66 33

**Classes **LowArray **and **LowArrayApp

In lowArray.java, we essentially wrap the class LowArray around an ordinary Java array. The array is hidden from the outside world inside the class; it’s private, so only LowArray class methods can access it. There are three LowArray methods: setElem() and getElem(), which insert and retrieve an element, respectively; and a constructor, which creates an empty array of a specified size.

Another class, LowArrayApp, creates an object of the LowArray class and uses it to store and manipulate data. Think of LowArray as a tool and LowArrayApp as a user of the tool. We’ve divided the program into two classes with clearly defined roles. This is a valuable first step in making a program object oriented.

A class used to store data objects, as is LowArray in the lowArray.java program, is sometimes called a

**Class Interfaces**

We’ve seen how a program can be divided into separate classes. How do these classes interact with each other? Communication between classes and the division of responsibility between them are important aspects of object-oriented programming.

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47

This point is especially true when a class may have many different users. Typically, a class can be used over and over by different users (or the same user) for different purposes. For example, someone might use the LowArray class in some other program to store the serial numbers of his traveler’s checks. The class can handle this task just as well as it can store the numbers of baseball players.

If a class is used by many different programmers, the class should be designed so that it’s easy to use. The way that a class user relates to the class is called the class

Private Data

tElem ()

a

se

getElem()

lowArray

Interface

**FIGURE 2.3**

The LowArray interface.

**Not So Convenient**

The interface to the LowArray class in lowArray.java is not particularly convenient.

The methods setElem() and getElem() operate on a low conceptual level, performing exactly the same tasks as the [] operator in an ordinary Java array. The class user, represented by the main() method in the LowArrayApp class, ends up having to carry out the same low-level operations it did in the non-class version of an array in the

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array.java program. The only difference was that it related to setElem() and getElem() instead of the [] operator. It’s not clear that this approach is an improvement.

Also notice that there’s no convenient way to display the contents of the array.

Somewhat crudely, the LowArrayApp class simply uses a for loop and the getElem() method for this purpose. We could avoid repeated code by writing a separate method for LowArrayApp that it could call to display the array contents, but is it really the responsibility of the LowArrayApp class to provide this method?

Thus, lowArray.java demonstrates how to divide a program into classes, but it really doesn’t buy us too much in practical terms. Let’s see how to redistribute responsibilities between the classes to obtain more of the advantages of OOP.

**Who’s Responsible for What? **

In the lowArray.java program, the main()routine in the LowArrayApp class, the user of the data storage structure, must keep track of the indices to the array. For some users of an array, who need random access to array elements and don’t mind keeping track of the index numbers, this arrangement might make sense. For example, sorting an array, as we’ll see in the next chapter, can make efficient use of this direct hands-on approach.

In a typical program, however, the user of the data storage device won’t find access to the array indices to be helpful or relevant.

**The **highArray.java **Example**

Out next example program shows an improved interface for the storage structure class, called HighArray. Using this interface, the class user (the HighArrayApp class) no longer needs to think about index numbers. The setElem() and getElem() methods are gone; they’re replaced by insert(), find(), and delete(). These new methods don’t require an index number as an argument because the class takes responsibility for handling index numbers. The user of the class (HighArrayApp) is free to concentrate on the

Figure 2.4 shows the HighArray interface, and Listing 2.3 shows the highArray.java program.

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49

Private Data

rt ()

inse

delete ()

a

Find()

highArray()

Interface

**FIGURE 2.4**

The HighArray interface.

**LISTING 2.3**

The highArray.java Program

// highArray.java

// demonstrates array class with high-level interface

// to run this program: C>java HighArrayApp

////////////////////////////////////////////////////////////////

class HighArray

{

private long[] a; // ref to array a private int nElems; // number of data items

//-----------------------------------------------------------

public HighArray(int max) // constructor

{

a = new long[max]; // create the array nElems = 0; // no items yet

}

//-----------------------------------------------------------

public boolean find(long searchKey)

{ // find specified value int j;

for(j=0; j<nElems; j++) // for each element, if(a[j] == searchKey) // found item?

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**LISTING 2.3**

Continued

break; // exit loop before end if(j == nElems) // gone to end?

return false; // yes, can’t find it else

return true; // no, found it

} // end find()

//-----------------------------------------------------------

public void insert(long value) // put element into array

{

a[nElems] = value; // insert it

nElems++; // increment size

}

//-----------------------------------------------------------

public boolean delete(long value)

{

int j;

for(j=0; j<nElems; j++) // look for it

if( value == a[j] )

break;

if(j==nElems) // can’t find it

return false;

else // found it

{

for(int k=j; k<nElems; k++) // move higher ones down a[k] = a[k+1];

nElems--; // decrement size

return true;

}

} // end delete()

//-----------------------------------------------------------

public void display() // displays array contents

{

for(int j=0; j<nElems; j++) // for each element, System.out.print(a[j] + “ “); // display it

System.out.println(“”);

}

//-----------------------------------------------------------

} // end class HighArray

////////////////////////////////////////////////////////////////

class HighArrayApp

{

public static void main(String[] args)

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**LISTING 2.3**

Continued

{

int maxSize = 100; // array size

HighArray arr; // reference to array arr = new HighArray(maxSize); // create the array

arr.insert(77); // insert 10 items

arr.insert(99);

arr.insert(44);

arr.insert(55);

arr.insert(22);

arr.insert(88);

arr.insert(11);

arr.insert(00);

arr.insert(66);

arr.insert(33);

arr.display(); // display items

int searchKey = 35; // search for item

if( arr.find(searchKey) )

System.out.println(“Found “ + searchKey);

else

System.out.println(“Can’t find “ + searchKey);

arr.delete(00); // delete 3 items

arr.delete(55);

arr.delete(99);

arr.display(); // display items again

} // end main()

} // end class HighArrayApp

////////////////////////////////////////////////////////////////

The HighArray class is now wrapped around the array. In main(), we create an array of this class and carry out almost the same operations as in the lowArray.java program: We insert 10 items, search for an item—one that isn’t there—and display the array contents. Because deleting is so easy, we delete 3 items (0, 55, and 99) instead of 1 and finally display the contents again. Here’s the output: 77 99 44 55 22 88 11 0 66 33

Can’t find 35

77 44 22 88 11 66 33

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Notice how short and simple main() is. The details that had to be handled by main() in lowArray.java are now handled by HighArray class methods.

In the HighArray class, the find() method looks through the array for the item whose key value was passed to it as an argument. It returns true or false, depending on whether it finds the item.

The insert() method places a new data item in the next available space in the array.

A field called nElems keeps track of the number of array cells that are actually filled with data items. The main() method no longer needs to worry about how many items are in the array.

The delete() method searches for the element whose key value was passed to it as an argument and, when it finds that element, shifts all the elements in higher index cells down one cell, thus writing over the deleted value; it then decrements nElems.

We’ve also included a display() method, which displays all the values stored in the array.

**The User’s Life Made Easier**

In lowArray.java (Listing 2.2), the code in main() to search for an item required eight lines; in highArray.java, it requires only one. The class user, the HighArrayApp class, need not worry about index numbers or any other array details. Amazingly, the class user doesn’t even need to know

**Abstraction**

The process of separating the

Abstraction is an important aspect of software engineering. By abstracting class functionality, we make it easier to design a program because we don’t need to think about implementation details at too early a stage in the design process.

**The Ordered Workshop Applet**

Imagine an array in which the data items are arranged in order of ascending key values—that is, with the smallest value at index 0, and each cell holding a value larger than the cell below. Such an array is called an

When we insert an item into this array, the correct location must be found for the insertion: just above a smaller value and just below a larger one. Then all the larger values must be moved up to make room.

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53

Why would we want to arrange data in order? One advantage is that we can speed up search times dramatically using a

Start the Ordered Workshop applet, using the procedure described in Chapter 1.

You’ll see an array; it’s similar to the one in the Array Workshop applet, but the data is ordered. Figure 2.5 shows this applet.

**FIGURE 2.5**

The Ordered Workshop applet.

In the ordered array we’ve chosen not to allow duplicates. As we saw earlier, this decision speeds up searching somewhat but slows down insertion.

**Linear Search**

Two search algorithms are available for the Ordered Workshop applet: linear and binary. Linear search is the default. Linear searches operate in much the same way as the searches in the unordered array in the Array applet: The red arrow steps along, looking for a match. The difference is that in the ordered array, the search quits if an item with a larger key is found.

Try out a linear search. Make sure the Linear radio button is selected. Then use the Find button to search for a non-existent value that, if it were present, would fit somewhere in the middle of the array. In Figure 2.5, this number might be 400.

You’ll see that the search terminates when the first item larger than 400 is reached; it’s 427 in the figure. The algorithm knows there’s no point looking further.

Try out the Ins and Del buttons as well. Use Ins to insert an item with a key value that will go somewhere in the middle of the existing items. You’ll see that insertion requires moving all the items with key values larger than the item being inserted.

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Use the Del button to delete an item from the middle of the array. Deletion works much the same as it did in the Array applet, shifting items with higher index numbers down to fill in the hole left by the deletion. In the ordered array, however, the deletion algorithm can quit partway through if it doesn’t find the item, just as the search routine can.

**Binary Search**

The payoff for using an ordered array comes when we use a binary search. This kind of search is much faster than a linear search, especially for large arrays.

**The Guess-a-Number Game**

Binary search uses the same approach you did as a kid (if you were smart) to guess a number in the well-known children’s guessing game. In this game, a friend asks you to guess a number she’s thinking of between 1 and 100. When you guess a number, she’ll tell you one of three things: Your guess is larger than the number she’s thinking of, it’s smaller, or you guessed correctly.

To find the number in the fewest guesses, you should always start by guessing 50. If your friend says your guess is too low, you deduce the number is between 51 and 100, so your next guess should be 75 (halfway between 51 and 100). If she says it’s too high, you deduce the number is between 1 and 49, so your next guess should be 25.

Each guess allows you to divide the range of possible values in half. Finally, the range is only one number long, and that’s the answer.

Notice how few guesses are required to find the number. If you used a linear search, guessing first 1, then 2, then 3, and so on, finding the number would take you, on the average, 50 guesses. In a binary search each guess divides the range of possible values in half, so the number of guesses required is far fewer. Table 2.2 shows a game session when the number to be guessed is 33.

**TABLE 2.2**

Guessing a Number

**Step Number**

**Number Guessed**

**Result**

**Range of Possible Values**

0

1–100

1

50

Too high

1–49

2

25

Too low

26–49

3

37

Too high

26–36

4

31

Too low

32–36

5

34

Too high

32–33

6

32

Too low

33–33

7

33

Correct

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55

The correct number is identified in only seven guesses. This is the maximum. You might get lucky and guess the number before you’ve worked your way all the way down to a range of one. This would happen if the number to be guessed was 50, for example, or 34.

**Binary Search in the Ordered Workshop Applet**

To perform a binary search with the Ordered Workshop applet, you must use the New button to create a new array. After the first press, you’ll be asked to specify the size of the array (maximum 60) and which kind of searching scheme you want: linear or binary. Choose binary by clicking the Binary radio button. After the array is created, use the Fill button to fill it with data items. When prompted, type the amount (not more than the size of the array). A few more presses fills in all the items.

When the array is filled, pick one of the values in the array and see how you can use the Find button to locate it. After a few preliminary presses, you’ll see the red arrow pointing to the algorithm’s current guess, and you’ll see the range shown by a vertical blue line adjacent to the appropriate cells. Figure 2.6 depicts the situation when the range is the entire array.

**FIGURE 2.6**

Initial range in the binary search.

At each press of the Find button, the range is halved and a new guess is chosen in the middle of the range. Figure 2.7 shows the next step in the process.

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**FIGURE 2.7**

Range in step 2 of the binary search.

Even with a maximum array size of 60 items, a half-dozen button presses suffices to locate any item.

Try using the binary search with different array sizes. Can you figure out how many steps are necessary before you run the applet? We’ll return to this question in the last section of this chapter.

Notice that the insertion and deletion operations also employ the binary search (when it’s selected). The place where an item should be inserted is found with a binary search, as is an item to be deleted. In this applet, items with duplicate keys are not permitted.

**Java Code for an Ordered Array**

Let’s examine some Java code that implements an ordered array. We’ll use the OrdArray class to encapsulate the array and its algorithms. The heart of this class is the find() method, which uses a binary search to locate a specified data item. We’ll examine this method in detail before showing the complete program.

**Binary Search with the **find() **Method**

The find() method searches for a specified item by repeatedly dividing in half the range of array elements to be considered. The method looks like this: public int find(long searchKey)

{

int lowerBound = 0;

int upperBound = nElems-1;

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57

int curIn;

while(true)

{

curIn = (lowerBound + upperBound ) / 2;

if(a[curIn]==searchKey)

return curIn; // found it

else if(lowerBound > upperBound)

return nElems; // can’t find it

else // divide range

{

if(a[curIn] < searchKey)

lowerBound = curIn + 1; // it’s in upper half

else

upperBound = curIn - 1; // it’s in lower half

} // end else divide range

} // end while

} // end find()

The method begins by setting the lowerBound and upperBound variables to the first and last occupied cells in the array. Setting these variables specifies the range where the item we’re looking for, searchKey, may be found. Then, within the while loop, the current index, curIn, is set to the middle of this range.

If we’re lucky, curIn may already be pointing to the desired item, so we first check if this is true. If it is, we’ve found the item, so we return with its index, curIn.

Each time through the loop we divide the range in half. Eventually, the range will get so small that it can’t be divided any more. We check for this in the next statement: If lowerBound is greater than upperBound, the range has ceased to exist. (When lowerBound equals upperBound, the range is one and we need one more pass through the loop.) We can’t continue the search without a valid range, but we haven’t found the desired item, so we return nElems, the total number of items. This isn’t a valid index because the last filled cell in the array is nElems-1. The class user interprets this value to mean that the item wasn’t found.

If curIn is not pointing at the desired item, and the range is still big enough, we’re ready to divide the range in half. We compare the value at the current index, a[curIn], which is in the middle of the range, with the value to be found, searchKey.

If searchKey is larger, we know we should look in the upper half of the range.

Accordingly, we move lowerBound up to curIn. Actually, we move it one cell beyond curIn because we’ve already checked curIn itself at the beginning of the loop.

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If searchKey is smaller than a[curIn], we know we should look in the lower half of the range. So we move upperBound down to one cell below curIn. Figure 2.8 shows how the range is altered in these two situations.

lowerBound

curIn

upperBound

lowerBound upperBound

lowerBound upperBound

curIn

curIn

New range if

New range if

searchKey<a[curIn]

searchKey>a[curIn]

**FIGURE 2.8**

Dividing the range in a binary search.

**The **OrdArray **Class**

In general, the orderedArray.java program is similar to highArray.java (Listing 2.3). The main difference is that find() uses a binary search, as we’ve seen.

We could have used a binary search to locate the position where a new item will be inserted. This operation involves a variation on the find() routine, but for simplicity we retain the linear search in insert(). The speed penalty may not be important because, as we’ve seen, an average of half the items must be moved anyway when an insertion is performed, so insertion will not be very fast even if we locate the item with a binary search. However, for the last ounce of speed, you could change the initial part of insert() to a binary search (as is done in the Ordered Workshop applet). Similarly, the delete() method could call find() to figure out the location of the item to be deleted.

The OrdArray class includes a new size() method, which returns the number of data items currently in the array. This information is helpful for the class user, main(), when it calls find(). If find() returns nElems, which main() can discover with size(), then the search was unsuccessful. Listing 2.4 shows the complete listing for the orderedArray.java program.

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59

**LISTING 2.4**

The orderedArray.java Program

// orderedArray.java

// demonstrates ordered array class

// to run this program: C>java OrderedApp

////////////////////////////////////////////////////////////////

class OrdArray

{

private long[] a; // ref to array a private int nElems; // number of data items

//-----------------------------------------------------------

public OrdArray(int max) // constructor

{

a = new long[max]; // create array

nElems = 0;

}

//-----------------------------------------------------------

public int size()

{ return nElems; }

//-----------------------------------------------------------

public int find(long searchKey)

{

int lowerBound = 0;

int upperBound = nElems-1;

int curIn;

while(true)

{

curIn = (lowerBound + upperBound ) / 2;

if(a[curIn]==searchKey)

return curIn; // found it

else if(lowerBound > upperBound)

return nElems; // can’t find it

else // divide range

{

if(a[curIn] < searchKey)

lowerBound = curIn + 1; // it’s in upper half

else

upperBound = curIn - 1; // it’s in lower half

} // end else divide range

} // end while

} // end find()

//-----------------------------------------------------------

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**LISTING 2.4**

Continued

public void insert(long value) // put element into array

{

int j;

for(j=0; j<nElems; j++) // find where it goes if(a[j] > value) // (linear search)

break;

for(int k=nElems; k>j; k--) // move bigger ones up a[k] = a[k-1];

a[j] = value; // insert it

nElems++; // increment size

} // end insert()

//-----------------------------------------------------------

public boolean delete(long value)

{

int j = find(value);

if(j==nElems) // can’t find it

return false;

else // found it

{

for(int k=j; k<nElems; k++) // move bigger ones down a[k] = a[k+1];

nElems--; // decrement size

return true;

}

} // end delete()

//-----------------------------------------------------------

public void display() // displays array contents

{

for(int j=0; j<nElems; j++) // for each element, System.out.print(a[j] + “ “); // display it

System.out.println(“”);

}

//-----------------------------------------------------------

} // end class OrdArray

////////////////////////////////////////////////////////////////

class OrderedApp

{

public static void main(String[] args)

{

int maxSize = 100; // array size

OrdArray arr; // reference to array

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**LISTING 2.4**

Continued

arr = new OrdArray(maxSize); // create the array

arr.insert(77); // insert 10 items

arr.insert(99);

arr.insert(44);

arr.insert(55);

arr.insert(22);

arr.insert(88);

arr.insert(11);

arr.insert(00);

arr.insert(66);

arr.insert(33);

int searchKey = 55; // search for item

if( arr.find(searchKey) != arr.size() )

System.out.println(“Found “ + searchKey);

else

System.out.println(“Can’t find “ + searchKey);

arr.display(); // display items

arr.delete(00); // delete 3 items

arr.delete(55);

arr.delete(99);

arr.display(); // display items again

} // end main()

} // end class OrderedApp

////////////////////////////////////////////////////////////////

**Advantages of Ordered Arrays**

What have we gained by using an ordered array? The major advantage is that search times are much faster than in an unordered array. The disadvantage is that insertion takes longer because all the data items with a higher key value must be moved up to make room. Deletions are slow in both ordered and unordered arrays because items must be moved down to fill the hole left by the deleted item.

Ordered arrays are therefore useful in situations in which searches are frequent, but insertions and deletions are not. An ordered array might be appropriate for a database of company employees, for example. Hiring new employees and laying off

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existing ones would probably be infrequent occurrences compared with accessing an existing employee’s record for information, or updating it to reflect changes in salary, address, and so on.

A retail store inventory, on the other hand, would not be a good candidate for an ordered array because the frequent insertions and deletions, as items arrived in the store and were sold, would run slowly.

**Logarithms**

In this section we’ll explain how logarithms are used to calculate the number of steps necessary in a binary search. If you’re a math major, you can probably skip this section. If math makes you break out in a rash, you can also skip it, except for taking a long hard look at Table 2.3.

We’ve seen that a binary search provides a significant speed increase over a linear search. In the number-guessing game, with a range from 1 to 100, a maximum of seven guesses is needed to identify any number using a binary search; just as in an array of 100 records, seven comparisons are needed to find a record with a specified key value. How about other ranges? Table 2.3 shows some representative ranges and the number of comparisons needed for a binary search.

**TABLE 2.3**

Comparisons Needed in Binary Search

**Range**

**Comparisons Needed**

10

4

100

7

1,000

10

10,000

14

100,000

17

1,000,000

20

10,000,000

24

100,000,000

27

1,000,000,000

30

Notice the differences between binary search times and linear search times. For very small numbers of items, the difference isn’t dramatic. Searching 10 items would take an average of five comparisons with a linear search (N/2) and a maximum of four comparisons with a binary search. But the more items there are, the bigger the difference. With 100 items, there are 50 comparisons in a linear search, but only 7 in a binary search. For 1,000 items, the numbers are 500 versus 10, and for 1,000,000

items, they’re 500,000 versus 20. We can conclude that for all but very small arrays, the binary search is greatly superior.

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63

**The Equation**

You can verify the results of Table 2.3 by repeatedly dividing a range (from the first column) in half until it’s too small to divide further. The number of divisions this process requires is the number of comparisons shown in the second column.

Repeatedly dividing the range by two is an algorithmic approach to finding the number of comparisons. You might wonder if you could also find the number using a simple equation. Of course, there is such an equation, and it’s worth exploring here because it pops up from time to time in the study of data structures. This formula involves logarithms. (Don’t panic yet.)

The numbers in Table 2.3 leave out some interesting data. They don’t answer such questions as, What is the exact size of the maximum range that can be searched in five steps? To solve this problem, we must create a similar table, but one that starts at the beginning, with a range of one, and works up from there by multiplying the range by two each time. Table 2.4 shows how this looks for the first seven steps.

**TABLE 2.4**

Powers of Two

**Step s, **

**Range r**

**Range Expressed**

**same as**

**as Power of**

**log (r)**

**2 (2s)**

**2**

0

1

20

1

2

21

2

4

22

3

8

23

4

16

24

5

32

25

6

64

26

7

128

27

8

256

28

9

512

29

10

1024

210

For our original problem with a range of 100, we can see that 6 steps don’t produce a range quite big enough (64), while 7 steps cover it handily (128). Thus, the 7 steps that are shown for 100 items in Table 2.3 are correct, as are the 10 steps for a range of 1000.

Doubling the range each time creates a series that’s the same as raising two to a power, as shown in the third column of Table 2.4. We can express this power as a formula. If s represents steps (the number of times you multiply by two—that is, the power to which two is raised) and r represents the range, then the equation is r = 2s

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If you know s, the number of steps, this tells you r, the range. For example, if s is 6, the range is 26, or 64.

**The Opposite of Raising Two to a Power**

Our original question was the opposite of the one just described: Given the range, we want to know how many comparisons are required to complete a search. That is, given r, we want an equation that gives us s.

The inverse of raising something to a power is called a logarithm. Here’s the formula we want, expressed with a logarithm:

s = log (r)

2

This equation says that the number of steps (comparisons) is equal to the logarithm to the base 2 of the range. What’s a logarithm? The base 2 logarithm of a number r is the number of times you must multiply two by itself to get r. In Table 2.4, we show that the numbers in the first column, s, are equal to log (r).

2

How do you find the logarithm of a number without doing a lot of dividing? Pocket calculators and most computer languages have a log function. It is usually log to the base 10, but you can convert easily to base 2 by multiplying by 3.322. For example, log (100) = 2, so log (100) = 2 times 3.322, or 6.644. Rounded up to the whole 10

2

number 7, this is what appears in the column to the right of 100 in Table 2.4.

In any case, the point here isn’t to calculate logarithms. It’s more important to understand the relationship between a number and its logarithm. Look again at Table 2.3, which compares the number of items and the number of steps needed to find a particular item. Every time you multiply the number of items (the range) by a factor of 10, you add only three or four steps (actually 3.322, before rounding off to whole numbers) to the number needed to find a particular element. This is true because, as a number grows larger, its logarithm doesn’t grow nearly as fast. We’ll compare this logarithmic growth rate with that of other mathematical functions when we talk about Big O notation later in this chapter.

**Storing Objects**

In the Java examples we’ve shown so far, we’ve stored primitive variables of type long in our data structures. Storing such variables simplifies the program examples, but it’s not representative of how you use data storage structures in the real world.

Usually, the data items (records) you want to store are combinations of many fields.

For a personnel record, you would store last name, first name, age, Social Security number, and so forth. For a stamp collection, you would store the name of the country that issued the stamp, its catalog number, condition, current value, and so on.

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In our next Java example, we’ll show how objects, rather than variables of primitive types, can be stored.

**The **Person **Class**

In Java, a data record is usually represented by a class object. Let’s examine a typical class used for storing personnel data. Here’s the code for the Person class: class Person

{

private String lastName;

private String firstName;

private int age;

//-----------------------------------------------------------

public Person(String last, String first, int a)

{ // constructor

lastName = last;

firstName = first;

age = a;

}

//-----------------------------------------------------------

public void displayPerson()

{

System.out.print(“ Last name: “ + lastName);

System.out.print(“, First name: “ + firstName);

System.out.println(“, Age: “ + age);

}

//-----------------------------------------------------------

public String getLast() // get last name

{ return lastName; }

} // end class Person

We show only three variables in this class, for a person’s last name, first name, and age. Of course, records for most applications would contain many additional fields.

A constructor enables a new Person object to be created and its fields initialized. The displayPerson() method displays a Person object’s data, and the getLast() method returns the Person’s last name; this is the key field used for searches.

**The **classDataArray.java **Program**

The program that makes use of the Person class is similar to the highArray.java program (Listing 2.3) that stored items of type long. Only a few changes are necessary to adapt that program to handle Person objects. Here are the major changes:

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• The type of the array a is changed to Person.

• The key field (the last name) is now a String object, so comparisons require the equals() method rather than the == operator. The getLast() method of Person obtains the last name of a Person object, and equals() does the comparison:

if( a[j].getLast().equals(searchName) ) // found item?

• The insert() method creates a new Person object and inserts it in the array, instead of inserting a long value.

The main() method has been modified slightly, mostly to handle the increased quantity of output. We still insert 10 items, display them, search for 1 item, delete 3

items, and display them all again. Listing 2.5 shows the complete classDataArray.java program.

**LISTING 2.5**

The classDataArray.java Program

// classDataArray.java

// data items as class objects

// to run this program: C>java ClassDataApp

////////////////////////////////////////////////////////////////

class Person

{

private String lastName;

private String firstName;

private int age;

//--------------------------------------------------------------

public Person(String last, String first, int a)

{ // constructor

lastName = last;

firstName = first;

age = a;

}

//--------------------------------------------------------------

public void displayPerson()

{

System.out.print(“ Last name: “ + lastName);

System.out.print(“, First name: “ + firstName);

System.out.println(“, Age: “ + age);

}

//--------------------------------------------------------------

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**LISTING 2.5**

Continued

public String getLast() // get last name

{ return lastName; }

} // end class Person

////////////////////////////////////////////////////////////////

class ClassDataArray

{

private Person[] a; // reference to array private int nElems; // number of data items public ClassDataArray(int max) // constructor

{

a = new Person[max]; // create the array nElems = 0; // no items yet

}

//--------------------------------------------------------------

public Person find(String searchName)

{ // find specified value int j;

for(j=0; j<nElems; j++) // for each element, if( a[j].getLast().equals(searchName) ) // found item?

break; // exit loop before end if(j == nElems) // gone to end?

return null; // yes, can’t find it else

return a[j]; // no, found it

} // end find()

//-------------------------------------------------------------

// put person into array

public void insert(String last, String first, int age)

{

a[nElems] = new Person(last, first, age);

nElems++; // increment size

}

//--------------------------------------------------------------

public boolean delete(String searchName)

{ // delete person from array int j;

for(j=0; j<nElems; j++) // look for it if( a[j].getLast().equals(searchName) )

break;

if(j==nElems) // can’t find it

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**LISTING 2.5**

Continued

return false;

else // found it

{

for(int k=j; k<nElems; k++) // shift down

a[k] = a[k+1];

nElems--; // decrement size

return true;

}

} // end delete()

//--------------------------------------------------------------

public void displayA() // displays array contents

{

for(int j=0; j<nElems; j++) // for each element, a[j].displayPerson(); // display it

}

//--------------------------------------------------------------

} // end class ClassDataArray

////////////////////////////////////////////////////////////////

class ClassDataApp

{

public static void main(String[] args)

{

int maxSize = 100; // array size

ClassDataArray arr; // reference to array arr = new ClassDataArray(maxSize); // create the array

// insert 10 items

arr.insert(“Evans”, “Patty”, 24);

arr.insert(“Smith”, “Lorraine”, 37);

arr.insert(“Yee”, “Tom”, 43);

arr.insert(“Adams”, “Henry”, 63);

arr.insert(“Hashimoto”, “Sato”, 21);

arr.insert(“Stimson”, “Henry”, 29);

arr.insert(“Velasquez”, “Jose”, 72);

arr.insert(“Lamarque”, “Henry”, 54);

arr.insert(“Vang”, “Minh”, 22);

arr.insert(“Creswell”, “Lucinda”, 18);

arr.displayA(); // display items

String searchKey = “Stimson”; // search for item

Person found;

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69

**LISTING 2.5**

Continued

found=arr.find(searchKey);

if(found != null)

{

System.out.print(“Found “);

found.displayPerson();

}

else

System.out.println(“Can’t find “ + searchKey);

System.out.println(“Deleting Smith, Yee, and Creswell”); arr.delete(“Smith”); // delete 3 items

arr.delete(“Yee”);

arr.delete(“Creswell”);

arr.displayA(); // display items again

} // end main()

} // end class ClassDataApp

////////////////////////////////////////////////////////////////

Here’s the output of this program:

Last name: Evans, First name: Patty, Age: 24

Last name: Smith, First name: Lorraine, Age: 37

Last name: Yee, First name: Tom, Age: 43

Last name: Adams, First name: Henry, Age: 63

Last name: Hashimoto, First name: Sato, Age: 21

Last name: Stimson, First name: Henry, Age: 29

Last name: Velasquez, First name: Jose, Age: 72

Last name: Lamarque, First name: Henry, Age: 54

Last name: Vang, First name: Minh, Age: 22

Last name: Creswell, First name: Lucinda, Age: 18

Found Last name: Stimson, First name: Henry, Age: 29

Deleting Smith, Yee, and Creswell

Last name: Evans, First name: Patty, Age: 24

Last name: Adams, First name: Henry, Age: 63

Last name: Hashimoto, First name: Sato, Age: 21

Last name: Stimson, First name: Henry, Age: 29

Last name: Velasquez, First name: Jose, Age: 72

Last name: Lamarque, First name: Henry, Age: 54

Last name: Vang, First name: Minh, Age: 22

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The classDataArray.java program shows that class objects can be handled by data storage structures in much the same way as primitive types. (Note that a serious program using the last name as a key would need to account for duplicate last names, which would complicate the programming as discussed earlier.) **Big O Notation **

Automobiles are divided by size into several categories: subcompacts, compacts, midsize, and so on. These categories provide a quick idea what size car you’re talking about, without needing to mention actual dimensions. Similarly, it’s useful to have a shorthand way to say how efficient a computer algorithm is. In computer science, this rough measure is called “Big O” notation.

You might think that in comparing algorithms you would say things like “Algorithm A is twice as fast as algorithm B,” but in fact this sort of statement isn’t too meaningful. Why not? Because the proportion can change radically as the number of items changes. Perhaps you increase the number of items by 50%, and now A is three times as fast as B. Or you have half as many items, and A and B are now equal. What you need is a comparison that tells how an algorithm’s speed is related to the number of items. Let’s see how this looks for the algorithms we’ve seen so far.

**Insertion in an Unordered Array: Constant**

Insertion into an unordered array is the only algorithm we’ve seen that doesn’t depend on how many items are in the array. The new item is always placed in the next available position, at a[nElems], and nElems is then incremented. Insertion requires the same amount of time no matter how big N—the number of items in the array—is. We can say that the time, T, to insert an item into an unsorted array is a constant K:

T = K

In a real situation, the actual time (in microseconds or whatever) required by the insertion is related to the speed of the microprocessor, how efficiently the compiler has generated the program code, and other factors. The constant K in the preceding equation is used to account for all such factors. To find out what K is in a real situation, you need to measure how long an insertion took. (Software exists for this very purpose.) K would then be equal to that time.

**Linear Search: Proportional to N**

We’ve seen that, in a linear search of items in an array, the number of comparisons that must be made to find a specified item is, on the average, half of the total number of items. Thus, if N is the total number of items, the search time T is proportional to half of N:

T = K * N / 2

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71

As with insertions, discovering the value of K in this equation would require timing a search for some (probably large) value of N and then using the resulting value of T

to calculate K. When you know K, you can calculate T for any other value of N.

For a handier formula, we could lump the 2 into the K. Our new K is equal to the old K divided by 2. Now we have

T = K * N

This equation says that average linear search times are proportional to the size of the array. If an array is twice as big, searching it will take twice as long.

**Binary Search: Proportional to log(N)**

Similarly, we can concoct a formula relating T and N for a binary search: T = K * log (N)

2

As we saw earlier, the time is proportional to the base 2 logarithm of N. Actually, because any logarithm is related to any other logarithm by a constant (3.322 to go from base 2 to base 10), we can lump this constant into K as well. Then we don’t need to specify the base:

T = K * log(N)

**Don’t Need the Constant**

Big O notation looks like the formulas just described, but it dispenses with the constant K. When comparing algorithms, you don’t really care about the particular microprocessor chip or compiler; all you want to compare is how T changes for different values of N, not what the actual numbers are. Therefore, the constant isn’t needed.

Big O notation uses the uppercase letter

“order of.” In Big O notation, we would say that a linear search takes O(N) time, and a binary search takes O(log N) time. Insertion into an unordered array takes O(1), or constant time. (That’s the numeral 1 in the parentheses.) Table 2.5 summarizes the running times of the algorithms we’ve discussed so far.

**TABLE 2.5**

Running Times in Big O Notation

**Algorithm**

**Running Time in Big O Notation**

Linear search

O(N)

Binary search

O(log N)

Insertion in unordered array

O(1)

Insertion in ordered array

O(N)

Deletion in unordered array

O(N)

Deletion in ordered array

O(N)

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Figure 2.9 graphs some Big O relationships between time and number of items. Based on this graph, we might rate the various Big O values (very subjectively) like this: O(1) is excellent, O(log N) is good, O(N) is fair, and O(N2) is poor. O(N2) occurs in the bubble sort and also in certain graph algorithms that we’ll look at later in this book.

40

35

30

O(N2)

25

20

Number of steps

15

O(N)

10

O(log N)

5

O(1)

0

5

10

15

20

25

Number of items (N)

**FIGURE 2.9**

Graph of Big O times.

The idea in Big O notation isn’t to give actual figures for running times but to convey how the running times are affected by the number of items. This is the most meaningful way to compare algorithms, except perhaps actually measuring running times in a real installation.

**Why Not Use Arrays for Everything? **

Arrays seem to get the job done, so why not use them for all data storage? We’ve already seen some of their disadvantages. In an unordered array you can insert items

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73

quickly, in O(1) time, but searching takes slow O(N) time. In an ordered array you can search quickly, in O(logN) time, but insertion takes O(N) time. For both kinds of arrays, deletion takes O(N) time because half the items (on the average) must be moved to fill in the hole.

It would be nice if there were data structures that could do everything—insertion, deletion, and searching—quickly, ideally in O(1) time, but if not that, then in O(logN) time. In the chapters ahead, we’ll see how closely this ideal can be approached, and the price that must be paid in complexity.

Another problem with arrays is that their size is fixed when they are first created with new. Usually, when the program first starts, you don’t know exactly how many items will be placed in the array later, so you guess how big it should be. If your guess is too large, you’ll waste memory by having cells in the array that are never filled. If your guess is too small, you’ll overflow the array, causing at best a message to the program’s user, and at worst a program crash.

Other data structures are more flexible and can expand to hold the number of items inserted in them. The linked list, discussed in Chapter 5, “Linked Lists,” is such a structure.

We should mention that Java includes a class called Vector that acts much like an array but is expandable. This added capability comes at the expense of some loss of efficiency.

You might want to try creating your own vector class. If the class user is about to overflow the internal array in this class, the insertion algorithm creates a new array of larger size, copies the old array contents to the new array, and then inserts the new item. This whole process would be invisible to the class user.

**Summary**

• Arrays in Java are objects, created with the new operator.

• Unordered arrays offer fast insertion but slow searching and deletion.

• Wrapping an array in a class protects the array from being inadvertently altered.

• A class interface is composed of the methods (and occasionally fields) that the class user can access.

• A class interface can be designed to make things simple for the class user.

• A binary search can be applied to an ordered array.

• The logarithm to the base B of a number A is (roughly) the number of times you can divide A by B before the result is less than 1.

• Linear searches require time proportional to the number of items in an array.

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• Binary searches require time proportional to the logarithm of the number of items.

• Big O notation provides a convenient way to compare the speed of algorithms.

• An algorithm that runs in O(1) time is the best, O(log N) is good, O(N) is fair, and O(N2) is pretty bad.

**Questions**

These questions are intended as a self-test for readers. Answers may be found in Appendix C.

**1. **Inserting an item into an unordered array **a. **takes time proportional to the size of the array.

**b. **requires multiple comparisons.

**c. **requires shifting other items to make room.

**d. **takes the same time no matter how many items there are.

**2. **True or False: When you delete an item from an unordered array, in most cases you shift other items to fill in the gap.

**3. **In an unordered array, allowing duplicates **a. **increases times for all operations.

**b. **increases search times in some situations.

**c. **always increases insertion times.

**d. **sometimes decreases insertion times.

**4. **True or False: In an unordered array, it’s generally faster to find out an item is not in the array than to find out it is.

**5. **Creating an array in Java requires using the keyword ________ .

**6. **If class A is going to use class B for something, then **a. **class A’s methods should be easy to understand.

**b. **it’s preferable if class B communicates with the program’s user.

**c. **the more complex operations should be placed in class A.

**d. **the more work that class B can do, the better.

**7. **When class A is using class B for something, the methods and fields class A can access in class B are called class B’s __________.

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75

**8. **Ordered arrays, compared with unordered arrays, are **a. **much quicker at deletion.

**b. **quicker at insertion.

**c. **quicker to create.

**d. **quicker at searching.

**9. **A logarithm is the inverse of _____________ .

**10. **The base 10 logarithm of 1,000 is _____ .

**11. **The maximum number of elements that must be examined to complete a binary search in an array of 200 elements is

**a. **200.

**b. **8.

**c. **1.

**d. **13.

**12. **The base 2 logarithm of 64 is ______ .

**13. **True or False: The base 2 logarithm of 100 is 2.

**14. **Big O notation tells

**a. **how the speed of an algorithm relates to the number of items.

**b. **the running time of an algorithm for a given size data structure.

**c. **the running time of an algorithm for a given number of items.

**d. **how the size of a data structure relates to the number of items.

**15. **O(1) means a process operates in _________ time.

**16. **Either variables of primitive types or _________ can be placed in an array.

**Experiments**

Carrying out these experiments will help to provide insights into the topics covered in the chapter. No programming is involved.

**1. **Use the Array Workshop applet to insert, search for, and delete items. Make sure you can predict what it’s going to do. Do this both when duplicates are allowed and when they’re not.

**2. **Make sure you can predict in advance what range the Ordered Workshop applet will select at each step.

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**3. **In an array holding an even number of data items, there is no middle item.

Which item does the binary search algorithm examine first? Use the Ordered Workshop applet to find out.

**Programming Projects**

Writing programs to solve the Programming Projects helps to solidify your understanding of the material and demonstrates how the chapter’s concepts are applied.

(As noted in the Introduction, qualified instructors may obtain completed solutions to the Programming Projects on the publisher’s Web site.) 2.1 To the HighArray class in the highArray.java program (Listing 2.3), add a method called getMax() that returns the value of the highest key in the array, or –1 if the array is empty. Add some code in main() to exercise this method.

You can assume all the keys are positive numbers.

2.2 Modify the method in Programming Project 2.1 so that the item with the highest key is not only returned by the method, but also removed from the array. Call the method removeMax().

2.3 The removeMax() method in Programming Project 2.2 suggests a way to sort the contents of an array by key value. Implement a sorting scheme that does not require modifying the HighArray class, but only the code in main(). You’ll need a second array, which will end up inversely sorted. (This scheme is a rather crude variant of the selection sort in Chapter 3, “Simple Sorting.”) 2.4 Modify the orderedArray.java program (Listing 2.4) so that the insert() and delete() routines, as well as find(), use a binary search, as suggested in the text.

2.5 Add a merge() method to the OrdArray class in the orderedArray.java program (Listing 2.4) so that you can merge two ordered source arrays into an ordered destination array. Write code in main() that inserts some random numbers into the two source arrays, invokes merge(), and displays the contents of the resulting destination array. The source arrays may hold different numbers of data items. In your algorithm you will need to compare the keys of the source arrays, picking the smallest one to copy to the destination. You’ll also need to handle the situation when one source array exhausts its contents before the other.

2.6 Write a noDups() method for the HighArray class of the highArray.java program (Listing 2.3). This method should remove all duplicates from the array. That is, if three items with the key 17 appear in the array, noDups() should remove two of them. Don’t worry about maintaining the order of the items. One approach is to first compare every item with all the other items and overwrite any duplicates with a null (or a distinctive value that isn’t used for real keys). Then remove all the nulls. Of course, the array size will be reduced.

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**3**

**IN THIS CHAPTER**

• How Would You Do It?

Simple Sorting

• Bubble Sort

• Selection Sort

• Insertion Sort

As soon as you create a significant database, you’ll probably think of reasons to sort it in various ways. You need to

• Sorting Objects

arrange names in alphabetical order, students by grade,

•

customers by ZIP code, home sales by price, cities in order Comparing the Simple Sorts

of increasing population, countries by GNP, stars by magnitude, and so on.

Sorting data may also be a preliminary step to searching it.

As we saw in Chapter 2, “Arrays,” a binary search, which can be applied only to sorted data, is much faster than a linear search.

Because sorting is so important and potentially so time-consuming, it has been the subject of extensive research in computer science, and some very sophisticated methods have been developed. In this chapter we’ll look at three of the simpler algorithms: the bubble sort, the selection sort, and the insertion sort. Each is demonstrated with its own Workshop applet. In Chapter 7, “Advanced Sorting,” we’ll look at more sophisticated approaches: Shellsort and quicksort.

The techniques described in this chapter, while unsophisticated and comparatively slow, are nevertheless worth examining. Besides being easier to understand, they are actually better in some circumstances than the more sophisticated algorithms. The insertion sort, for example, is preferable to quicksort for small files and for almost-sorted files. In fact, an insertion sort is commonly used as a part of a quicksort implementation.

The example programs in this chapter build on the array classes we developed in the preceding chapter. The sorting algorithms are implemented as methods of similar array classes.

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Be sure to try out the Workshop applets included in this chapter. They are more effective in explaining how the sorting algorithms work than prose and static pictures could ever be.

**How Would You Do It? **

Imagine that your kids-league baseball team (mentioned in Chapter 1, “Overview”) is lined up on the field, as shown in Figure 3.1. The regulation nine players, plus an extra, have shown up for practice. You want to arrange the players in order of increasing height (with the shortest player on the left) for the team picture. How would you go about this sorting process?

**FIGURE 3.1**

The unordered baseball team.

As a human being, you have advantages over a computer program. You can see all the kids at once, and you can pick out the tallest kid almost instantly. You don’t need to laboriously measure and compare everyone. Also, the kids don’t need to occupy particular places. They can jostle each other, push each other a little to make room, and stand behind or in front of each other. After some ad hoc rearranging, you would have no trouble in lining up all the kids, as shown in Figure 3.2.

**FIGURE 3.2**

The ordered baseball team.

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79

A computer program isn’t able to glance over the data in this way. It can compare only two players at one time because that’s how the comparison operators work. This tunnel vision on the part of algorithms will be a recurring theme. Things may seem simple to us humans, but the algorithm can’t see the big picture and must, therefore, concentrate on the details and follow some simple rules.

The three algorithms in this chapter all involve two steps, executed over and over until the data is sorted:

**1. **Compare two items.

**2. **Swap two items, or copy one item.

However, each algorithm handles the details in a different way.

**Bubble Sort**

The bubble sort is notoriously slow, but it’s conceptually the simplest of the sorting algorithms and for that reason is a good beginning for our exploration of sorting techniques.

**Bubble Sort on the Baseball Players**

Imagine that you’re near-sighted (like a computer program) so that you can see only two of the baseball players at the same time, if they’re next to each other and if you stand very close to them. Given this impediment, how would you sort them? Let’s assume there are N players, and the positions they’re standing in are numbered from 0 on the left to N-1 on the right.

The bubble sort routine works like this: You start at the left end of the line and compare the two kids in positions 0 and 1. If the one on the left (in 0) is taller, you swap them. If the one on the right is taller, you don’t do anything. Then you move over one position and compare the kids in positions 1 and 2. Again, if the one on the left is taller, you swap them. This sorting process is shown in Figure 3.3.

Here are the rules you’re following:

**1. **Compare two players.

**2. **If the one on the left is taller, swap them.

**3. **Move one position right.

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Swap

0

1

2

3

4

5

6

7

8

9

No Swap

Swap

Swap

**FIGURE 3.3**

Bubble sort: the beginning of the first pass.

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You continue down the line this way until you reach the right end. You have by no means finished sorting the kids, but you do know that the tallest kid is on the right. This must be true because, as soon as you encounter the tallest kid, you’ll end up swapping him (or her) every time you compare two kids, until eventually he (or she) will reach the right end of the line. This is why it’s called the bubble sort: As the algorithm progresses, the biggest items “bubble up” to the top end of the array. Figure 3.4 shows the baseball players at the end of the first pass.

Sorted

**FIGURE 3.4**

Bubble sort: the end of the first pass.

After this first pass through all the data, you’ve made N-1 comparisons and somewhere between 0 and N-1 swaps, depending on the initial arrangement of the players. The item at the end of the array is sorted and won’t be moved again.

Now you go back and start another pass from the left end of the line. Again, you go toward the right, comparing and swapping when appropriate. However, this time you can stop one player short of the end of the line, at position N-2, because you know the last position, at N-1, already contains the tallest player.

This rule could be stated as:

**4. **When you reach the first sorted player, start over at the left end of the line.

You continue this process until all the players are in order. Describing this process is much harder than demonstrating it, so let’s watch the BubbleSort Workshop applet at work.

**The BubbleSort Workshop Applet**

Start the BubbleSort Workshop applet. You’ll see something that looks like a bar graph, with the bar heights randomly arranged, as shown in Figure 3.5.

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**FIGURE 3.5**

The BubbleSort Workshop applet.

**The Run Button**

This Workshop applet contains a two-speed graph: You can either let it run by itself, or you can single-step through the process. To get a quick idea what happens, click the Run button. The algorithm will bubble-sort the bars. When it finishes, in 10

seconds or so, the bars will be sorted, as shown in Figure 3.6.

**FIGURE 3.6**

After the bubble sort.

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**The New Button**

To do another sort, press the New button. New creates a new set of bars and initializes the sorting routine. Repeated presses of New toggle between two arrangements of bars: a random order, as shown in Figure 3.5, and an inverse ordering where the bars are sorted backward. This inverse ordering provides an extra challenge for many sorting algorithms.

**The Step Button**

The real payoff for using the BubbleSort Workshop applet comes when you single-step through a sort. You can see exactly how the algorithm carries out each step.

Start by creating a new randomly arranged graph with New. You’ll see three arrows pointing at different bars. Two arrows, labeled inner and inner+1, are side by side on the left. Another arrow, outer, starts on the far right. (The names are chosen to correspond to the inner and outer loop variables in the nested loops used in the algorithm.)

Click once on the Step button. You’ll see the inner and the inner+1 arrows move together one position to the right, swapping the bars if appropriate. These arrows correspond to the two players you compared, and possibly swapped, in the baseball scenario.

A message under the arrows tells you whether the contents of inner and inner+1 will be swapped, but you know this just from comparing the bars: If the taller one is on the left, they’ll be swapped. Messages at the top of the graph tell you how many swaps and comparisons have been carried out so far. (A complete sort of 10 bars requires 45 comparisons and, on the average, about 22 swaps.) Continue pressing Step. Each time inner and inner+1 finish going all the way from 0

to outer, the outer pointer moves one position to the left. At all times during the sorting process, all the bars to the right of outer are sorted; those to the left of (and at) outer are not.

**The Size Button**

The Size button toggles between 10 bars and 100 bars. Figure 3.7 shows what the 100

random bars look like.

You probably don’t want to single-step through the sorting process for 100 bars, unless you’re unusually patient. Press Run instead, and watch how the blue inner and inner+1 pointers seem to find the tallest unsorted bar and carry it down the row to the right, inserting it just to the left of the previously sorted bars.

Figure 3.8 shows the situation partway through the sorting process. The bars to the right of the red (longest) arrow are sorted. The bars to the left are beginning to look sorted, but much work remains to be done.

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**FIGURE 3.7**

The BubbleSort applet with 100 bars.

**FIGURE 3.8**

The 100 partly sorted bars.

If you started a sort with Run and the arrows are whizzing around, you can freeze the process at any point by pressing the Step button. You can then single-step to watch the details of the operation or press Run again to return to high-speed mode.

**The Draw Button**

Sometimes while running the sorting algorithm at full speed, the computer takes time off to perform some other task. This can result in some bars not being drawn. If this happens, you can press the Draw button to redraw all the bars. Doing so pauses the run, so you’ll need to press the Run button again to continue.

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You can press Draw at any time there seems to be a glitch in the display.

**Java Code for a Bubble Sort**

In the bubbleSort.java program, shown in Listing 3.1, a class called ArrayBub encapsulates an array a[], which holds variables of type long.

In a more serious program, the data would probably consist of objects, but we use a primitive type for simplicity. (We’ll see how objects are sorted in the objectSort.java program in Listing 3.4.) Also, to reduce the size of the listing, we don’t show find() and delete() methods with the ArrayBub class, although they would normally be part of a such a class.

**LISTING 3.1**

The bubbleSort.java Program

// bubbleSort.java

// demonstrates bubble sort

// to run this program: C>java BubbleSortApp

////////////////////////////////////////////////////////////////

class ArrayBub

{

private long[] a; // ref to array a private int nElems; // number of data items

//--------------------------------------------------------------

public ArrayBub(int max) // constructor

{

a = new long[max]; // create the array nElems = 0; // no items yet

}

//--------------------------------------------------------------

public void insert(long value) // put element into array

{

a[nElems] = value; // insert it

nElems++; // increment size

}

//--------------------------------------------------------------

public void display() // displays array contents

{

for(int j=0; j<nElems; j++) // for each element, System.out.print(a[j] + “ “); // display it

System.out.println(“”);

}

//--------------------------------------------------------------

public void bubbleSort()

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**LISTING 3.1**

Continued

{

int out, in;

for(out=nElems-1; out>1; out--) // outer loop (backward) for(in=0; in<out; in++) // inner loop (forward) if( a[in] > a[in+1] ) // out of order?

swap(in, in+1); // swap them

} // end bubbleSort()

//--------------------------------------------------------------

private void swap(int one, int two)

{

long temp = a[one];

a[one] = a[two];

a[two] = temp;

}

//--------------------------------------------------------------

} // end class ArrayBub

////////////////////////////////////////////////////////////////

class BubbleSortApp

{

public static void main(String[] args)

{

int maxSize = 100; // array size

ArrayBub arr; // reference to array arr = new ArrayBub(maxSize); // create the array

arr.insert(77); // insert 10 items

arr.insert(99);

arr.insert(44);

arr.insert(55);

arr.insert(22);

arr.insert(88);

arr.insert(11);

arr.insert(00);

arr.insert(66);

arr.insert(33);

arr.display(); // display items

arr.bubbleSort(); // bubble sort them

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Bubble Sort

87

**LISTING 3.1**

Continued

arr.display(); // display them again

} // end main()

} // end class BubbleSortApp

////////////////////////////////////////////////////////////////

The constructor and the insert() and display() methods of this class are similar to those we’ve seen before. However, there’s a new method: bubbleSort(). When this method is invoked from main(), the contents of the array are rearranged into sorted order.

The main() routine inserts 10 items into the array in random order, displays the array, calls bubbleSort() to sort it, and then displays it again. Here’s the output: 77 99 44 55 22 88 11 0 66 33

0 11 22 33 44 55 66 77 88 99

The bubbleSort() method is only four lines long. Here it is, extracted from the listing: public void bubbleSort()

{

int out, in;

for(out=nElems-1; out>1; out--) // outer loop (backward) for(in=0; in<out; in++) // inner loop (forward) if( a[in] > a[in+1] ) // out of order?

swap(in, in+1); // swap them

} // end bubbleSort()

The idea is to put the smallest item at the beginning of the array (index 0) and the largest item at the end (index nElems-1). The loop counter out in the outer for loop starts at the end of the array, at nElems-1, and decrements itself each time through the loop. The items at indices greater than out are always completely sorted. The out variable moves left after each pass by in so that items that are already sorted are no longer involved in the algorithm.

The inner loop counter in starts at the beginning of the array and increments itself each cycle of the inner loop, exiting when it reaches out. Within the inner loop, the two array cells pointed to by in and in+1 are compared, and swapped if the one in in is larger than the one in in+1.

For clarity, we use a separate swap() method to carry out the swap. It simply exchanges the two values in the two array cells, using a temporary variable to hold the value of the first cell while the first cell takes on the value in the second and

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then setting the second cell to the temporary value. Actually, using a separate swap() method may not be a good idea in practice because the function call adds a small amount of overhead. If you’re writing your own sorting routine, you may prefer to put the swap instructions in line to gain a slight increase in speed.

**Invariants**

In many algorithms there are conditions that remain unchanged as the algorithm proceeds. These conditions are called

In the bubbleSort.java program, the invariant is that the data items to the right of out are sorted. This remains true throughout the running of the algorithm. (On the first pass, nothing has been sorted yet, and there are no items to the right of out because it starts on the rightmost element.)

**Efficiency of the Bubble Sort**

As you can see by watching the BubbleSort Workshop applet with 10 bars, the inner and inner+1 arrows make nine comparisons on the first pass, eight on the second, and so on, down to one comparison on the last pass. For 10 items, this is 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 45

In general, where N is the number of items in the array, there are N-1 comparisons on the first pass, N-2 on the second, and so on. The formula for the sum of such a series is

(N–1) + (N–2) + (N–3) + ... + 1 = N*(N–1)/2

N*(N–1)/2 is 45 (10*9/2) when N is 10.

Thus, the algorithm makes about N2⁄2 comparisons (ignoring the –1, which doesn’t make much difference, especially if N is large).

There are fewer swaps than there are comparisons because two bars are swapped only if they need to be. If the data is random, a swap is necessary about half the time, so there will be about N2⁄4 swaps. (Although in the worst case, with the initial data inversely sorted, a swap is necessary with every comparison.) Both swaps and comparisons are proportional to N2. Because constants don’t count in Big O notation, we can ignore the 2 and the 4 and say that the bubble sort runs in O(N2) time. This is slow, as you can verify by running the BubbleSort Workshop applet with 100 bars.

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Whenever you see one loop nested within another, such as those in the bubble sort and the other sorting algorithms in this chapter, you can suspect that an algorithm runs in O(N2) time. The outer loop executes N times, and the inner loop executes N

(or perhaps N divided by some constant) times for each cycle of the outer loop. This means you’re doing something approximately N*N or N2 times.

**Selection Sort**

The selection sort improves on the bubble sort by reducing the number of swaps necessary from O(N2) to O(N). Unfortunately, the number of comparisons remains O(N2). However, the selection sort can still offer a significant improvement for large records that must be physically moved around in memory, causing the swap time to be much more important than the comparison time. (Typically, this isn’t the case in Java, where references are moved around, not entire objects.) **Selection Sort on the Baseball Players**

Let’s consider the baseball players again. In the selection sort, you can no longer compare only players standing next to each other. Thus, you’ll need to remember a certain player’s height; you can use a notebook to write it down. A magenta-colored towel will also come in handy.

**A Brief Description**

What’s involved in the selection sort is making a pass through all the players and picking (or

The next time you pass down the row of players, you start at position 1, and, finding the minimum, swap with position 1. This process continues until all the players are sorted.

**A More Detailed Description**

In more detail, start at the left end of the line of players. Record the leftmost player’s height in your notebook and throw the magenta towel on the ground in front of this person. Then compare the height of the next player to the right with the height in your notebook. If this player is shorter, cross out the height of the first player and record the second player’s height instead. Also move the towel, placing it in front of this new “shortest” (for the time being) player. Continue down the row, comparing each player with the minimum. Change the minimum value in your notebook and move the towel whenever you find a shorter player. When you’re done, the magenta towel will be in front of the shortest player.

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Swap this shortest player with the player on the left end of the line. You’ve now sorted one player. You’ve made N-1 comparisons, but only one swap.

On the next pass, you do exactly the same thing, except that you can completely ignore the player on the left because this player has already been sorted. Thus, the algorithm starts the second pass at position 1, instead of 0. With each succeeding pass, one more player is sorted and placed on the left, and one less player needs to be considered when finding the new minimum. Figure 3.9 shows how this sort looks for the first three passes.

**The SelectSort Workshop Applet**

To see how the selection sort looks in action, try out the SelectSort Workshop applet.

The buttons operate the same way as those in the BubbleSort applet. Use New to create a new array of 10 randomly arranged bars. The red arrow called outer starts on the left; it points to the leftmost unsorted bar. Gradually, it will move right as more bars are added to the sorted group on its left.

The magenta min arrow also starts out pointing to the leftmost bar; it will move to record the shortest bar found so far. (The magenta min arrow corresponds to the towel in the baseball analogy.) The blue inner arrow marks the bar currently being compared with the minimum.

As you repeatedly press Step, inner moves from left to right, examining each bar in turn and comparing it with the bar pointed to by min. If the inner bar is shorter, min jumps over to this new, shorter bar. When inner reaches the right end of the graph, min points to the shortest of the unsorted bars. This bar is then swapped with outer, the leftmost unsorted bar.

Figure 3.10 shows the situation midway through a sort. The bars to the left of outer are sorted, and inner has scanned from outer to the right end, looking for the shortest bar. The min arrow has recorded the position of this bar, which will be swapped with outer.

Use the Size button to switch to 100 bars, and sort a random arrangement. You’ll see how the magenta min arrow hangs out with a perspective minimum value for a while and then jumps to a new one when the blue inner arrow finds a smaller candidate.

The red outer arrow moves slowly but inexorably to the right, as the sorted bars accumulate to its left.

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91

Swap

Minimum

Minimum

(No Swap)

Sorted

Swap

Minimum

Sorted

Swap

Minimum

Sorted

**FIGURE 3.9**

Selection sort on baseball players.

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**FIGURE 3.10**

The SelectSort Workshop applet.

**Java Code for Selection Sort**

The listing for the selectSort.java program is similar to that for bubbleSort.java, except that the container class is called ArraySel instead of ArrayBub, and the bubbleSort() method has been replaced by selectSort(). Here’s how this method looks:

public void selectionSort()

{

int out, in, min;

for(out=0; out<nElems-1; out++) // outer loop

{

min = out; // minimum

for(in=out+1; in<nElems; in++) // inner loop

if(a[in] < a[min] ) // if min greater,

min = in; // we have a new min

swap(out, min); // swap them

} // end for(out)

} // end selectionSort()

The outer loop, with loop variable out, starts at the beginning of the array (index 0) and proceeds toward higher indices. The inner loop, with loop variable in, begins at out and likewise proceeds to the right.

At each new position of in, the elements a[in] and a[min] are compared. If a[in] is smaller, then min is given the value of in. At the end of the inner loop, min points to

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the minimum value, and the array elements pointed to by out and min are swapped.

Listing 3.2 shows the complete selectSort.java program.

**LISTING 3.2**

The selectSort.java Program

// selectSort.java

// demonstrates selection sort

// to run this program: C>java SelectSortApp

////////////////////////////////////////////////////////////////

class ArraySel

{

private long[] a; // ref to array a private int nElems; // number of data items

//--------------------------------------------------------------

public ArraySel(int max) // constructor

{

a = new long[max]; // create the array nElems = 0; // no items yet

}

//--------------------------------------------------------------

public void insert(long value) // put element into array

{

a[nElems] = value; // insert it

nElems++; // increment size

}

//--------------------------------------------------------------

public void display() // displays array contents

{

for(int j=0; j<nElems; j++) // for each element, System.out.print(a[j] + “ “); // display it

System.out.println(“”);

}

//--------------------------------------------------------------

public void selectionSort()

{

int out, in, min;

for(out=0; out<nElems-1; out++) // outer loop

{

min = out; // minimum

for(in=out+1; in<nElems; in++) // inner loop

if(a[in] < a[min] ) // if min greater,

min = in; // we have a new min

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**LISTING 3.2**

Continued

swap(out, min); // swap them

} // end for(out)

} // end selectionSort()

//--------------------------------------------------------------

private void swap(int one, int two)

{

long temp = a[one];

a[one] = a[two];

a[two] = temp;

}

//--------------------------------------------------------------

} // end class ArraySel

////////////////////////////////////////////////////////////////

class SelectSortApp

{

public static void main(String[] args)

{

int maxSize = 100; // array size

ArraySel arr; // reference to array arr = new ArraySel(maxSize); // create the array

arr.insert(77); // insert 10 items

arr.insert(99);

arr.insert(44);

arr.insert(55);

arr.insert(22);

arr.insert(88);

arr.insert(11);

arr.insert(00);

arr.insert(66);

arr.insert(33);

arr.display(); // display items

arr.selectionSort(); // selection-sort them arr.display(); // display them again

} // end main()

} // end class SelectSortApp

////////////////////////////////////////////////////////////////

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The output from selectSort.java is identical to that from bubbleSort.java: 77 99 44 55 22 88 11 0 66 33

0 11 22 33 44 55 66 77 88 99

**Invariant**

In the selectSort.java program, the data items with indices less than or equal to out are always sorted.

**Efficiency of the Selection Sort**

The selection sort performs the same number of comparisons as the bubble sort: N*(N-1)/2. For 10 data items, this is 45 comparisons. However, 10 items require fewer than 10 swaps. With 100 items, 4,950 comparisons are required, but fewer than 100

swaps. For large values of N, the comparison times will dominate, so we would have to say that the selection sort runs in O(N2) time, just as the bubble sort did. However, it is unquestionably faster because there are so few swaps. For smaller values of N, the selection sort may in fact be considerably faster, especially if the swap times are much larger than the comparison times.

**Insertion Sort**

In most cases the insertion sort is the best of the elementary sorts described in this chapter. It still executes in O(N2) time, but it’s about twice as fast as the bubble sort and somewhat faster than the selection sort in normal situations. It’s also not too complex, although it’s slightly more involved than the bubble and selection sorts.

It’s often used as the final stage of more sophisticated sorts, such as quicksort.

**Insertion Sort on the Baseball Players**

To begin the insertion sort, start with your baseball players lined up in random order.

(They wanted to play a game, but clearly there’s no time for that.) It’s easier to think about the insertion sort if we begin in the middle of the process, when the team is half sorted.

**Partial Sorting**

At this point there’s an imaginary marker somewhere in the middle of the line.

(Maybe you threw a red T-shirt on the ground in front of a player.) The players to the left of this marker are

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Note that partial sorting did not take place in the bubble sort and selection sort. In these algorithms a group of data items was completely sorted at any given time; in the insertion sort a group of items is only partially sorted.

**The Marked Player**

The player where the marker is, whom we’ll call the “marked” player, and all the players on her right, are as yet unsorted. This is shown in Figure 3.11.a.

a)

Partially

"Marked" player

Sorted

b)

Empty space

To be shifted

(Taller than marked player)

c)

Inserted

"Marked" player

Shifted

Internally sorted

**FIGURE 3.11**

The insertion sort on baseball players.

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What we’re going to do is insert the marked player in the appropriate place in the (partially) sorted group. However, to do this, we’ll need to shift some of the sorted players to the right to make room. To provide a space for this shift, we take the marked player out of line. (In the program this data item is stored in a temporary variable.) This step is shown in Figure 3.11.b.

Now we shift the sorted players to make room. The tallest sorted player moves into the marked player’s spot, the next-tallest player into the tallest player’s spot, and so on.

When does this shifting process stop? Imagine that you and the marked player are walking down the line to the left. At each position you shift another player to the right, but you also compare the marked player with the player about to be shifted.

The shifting process stops when you’ve shifted the last player that’s taller than the marked player. The last shift opens up the space where the marked player, when inserted, will be in sorted order. This step is shown in Figure 3.11.c.

Now the partially sorted group is one player bigger, and the unsorted group is one player smaller. The marker T-shirt is moved one space to the right, so it’s again in front of the leftmost unsorted player. This process is repeated until all the unsorted players have been inserted (hence the name

**The InsertSort Workshop Applet**

Use the InsertSort Workshop applet to demonstrate the insertion sort. Unlike the other sorting applets, it’s probably more instructive to begin with 100 random bars rather than 10.

**Sorting 100 Bars**

Change to 100 bars with the Size button, and click Run to watch the bars sort themselves before your very eyes. You’ll see that the short red outer arrow marks the dividing line between the partially sorted bars to the left and the unsorted bars to the right. The blue inner arrow keeps starting from outer and zipping to the left, looking for the proper place to insert the marked bar. Figure 3.12 shows how this process looks when about half the bars are partially sorted.

The marked bar is stored in the temporary variable pointed to by the magenta arrow at the right end of the graph, but the contents of this variable are replaced so often that it’s hard to see what’s there (unless you slow down to single-step mode).

**Sorting 10 Bars**

To get down to the details, use Size to switch to 10 bars. (If necessary, use New to make sure they’re in random order.)

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**FIGURE 3.12**

The InsertSort Workshop applet with 100 bars.

At the beginning, inner and outer point to the second bar from the left (array index 1), and the first message is Will copy outer to temp. This will make room for the shift. (There’s no arrow for inner-1, but of course it’s always one bar to the left of inner.)

Click the Step button. The bar at outer will be copied to temp. We say that items are copied from a source to a destination. When performing a copy, the applet removes the bar from the source location, leaving a blank. This is slightly misleading because in a real Java program the reference in the source would remain there. However, blanking the source makes it easier to see what’s happening.

What happens next depends on whether the first two bars are already in order (smaller on the left). If they are, you’ll see the message Have compared inner-1 and temp, no copy necessary.

If the first two bars are not in order, the message is Have compared inner-1 and temp, will copy inner-1 to inner. This is the shift that’s necessary to make room for the value in temp to be reinserted. There’s only one such shift on this first pass; more shifts will be necessary on subsequent passes. The situation is shown in Figure 3.13.

On the next click, you’ll see the copy take place from inner-1 to inner. Also, the inner arrow moves one space left. The new message is Now inner is 0, so no copy necessary. The shifting process is complete.

No matter which of the first two bars was shorter, the next click will show you Will copy temp to inner. This will happen, but if the first two bars were initially in order, you won’t be able to tell a copy was performed because temp and inner hold the same bar. Copying data over the top of the same data may seem inefficient, but the algorithm runs faster if it doesn’t check for this possibility, which happens comparatively infrequently.

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**FIGURE 3.13**

The InsertSort Workshop applet with 10 bars.

Now the first two bars are partially sorted (sorted with respect to each other), and the outer arrow moves one space right, to the third bar (index 2). The process repeats, with the Will copy outer to temp message. On this pass through the sorted data, there may be no shifts, one shift, or two shifts, depending on where the third bar fits among the first two.

Continue to single-step the sorting process. Again, you can see what’s happening more easily after the process has run long enough to provide some sorted bars on the left. Then you can see how just enough shifts take place to make room for the reinsertion of the bar from temp into its proper place.

**Java Code for Insertion Sort**

Here’s the method that carries out the insertion sort, extracted from the insertSort.java program:

public void insertionSort()

{

int in, out;

for(out=1; out<nElems; out++) // out is dividing line

{

long temp = a[out]; // remove marked item

in = out; // start shifts at out while(in>0 && a[in-1] >= temp) // until one is smaller,

{

a[in] = a[in-1]; // shift item right,

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--in; // go left one position

}

a[in] = temp; // insert marked item

} // end for

} // end insertionSort()

In the outer for loop, out starts at 1 and moves right. It marks the leftmost unsorted data. In the inner while loop, in starts at out and moves left, until either temp is smaller than the array element there, or it can’t go left any further. Each pass through the while loop shifts another sorted element one space right.

It may be hard to see the relation between the steps in the InsertSort Workshop applet and the code, so Figure 3.14 is an activity diagram of the insertionSort() method, with the corresponding messages from the InsertSort Workshop applet.

Listing 3.3 shows the complete insertSort.java program.

outer=1

[outer==nElems]

[else]

“Will copy

outer to temp”

temp=a[outer]

inner=outer

“Will copy

temp to inner”

[else]

[inner>0]

a[inner]=temp

outer++

“Have compared

inner–1 and temp.

No copy necessary”

[else]

“Have compared

a[inner–1]

inner–1 and temp.

>=temp

Will copy inner–1

to inner”

a[inner]=a[inner–1]

--inner

**FIGURE 3.14**

Activity diagram for insertSort().

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101

**LISTING 3.3**

The insertSort.java Program

// insertSort.java

// demonstrates insertion sort

// to run this program: C>java InsertSortApp

//--------------------------------------------------------------

class ArrayIns

{

private long[] a; // ref to array a private int nElems; // number of data items

//--------------------------------------------------------------

public ArrayIns(int max) // constructor

{

a = new long[max]; // create the array nElems = 0; // no items yet

}

//--------------------------------------------------------------

public void insert(long value) // put element into array

{

a[nElems] = value; // insert it

nElems++; // increment size

}

//--------------------------------------------------------------

public void display() // displays array contents

{

for(int j=0; j<nElems; j++) // for each element, System.out.print(a[j] + “ “); // display it

System.out.println(“”);

}

//--------------------------------------------------------------

public void insertionSort()

{

int in, out;

for(out=1; out<nElems; out++) // out is dividing line

{

long temp = a[out]; // remove marked item in = out; // start shifts at out while(in>0 && a[in-1] >= temp) // until one is smaller,

{

a[in] = a[in-1]; // shift item to right

--in; // go left one position

}

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**LISTING 3.3**

Continued

a[in] = temp; // insert marked item

} // end for

} // end insertionSort()

//--------------------------------------------------------------

} // end class ArrayIns

////////////////////////////////////////////////////////////////

class InsertSortApp

{

public static void main(String[] args)

{

int maxSize = 100; // array size

ArrayIns arr; // reference to array arr = new ArrayIns(maxSize); // create the array

arr.insert(77); // insert 10 items

arr.insert(99);

arr.insert(44);

arr.insert(55);

arr.insert(22);

arr.insert(88);

arr.insert(11);

arr.insert(00);

arr.insert(66);

arr.insert(33);

arr.display(); // display items

arr.insertionSort(); // insertion-sort them arr.display(); // display them again

} // end main()

} // end class InsertSortApp

////////////////////////////////////////////////////////////////

Here’s the output from the insertSort.java program; it’s the same as that from the other programs in this chapter:

77 99 44 55 22 88 11 0 66 33

0 11 22 33 44 55 66 77 88 99

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103

**Invariants in the Insertion Sort**

At the end of each pass, following the insertion of the item from temp, the data items with smaller indices than outer are partially sorted.

**Efficiency of the Insertion Sort**

How many comparisons and copies does this algorithm require? On the first pass, it compares a maximum of one item. On the second pass, it’s a maximum of two items, and so on, up to a maximum of N-1 comparisons on the last pass. This is 1 + 2 + 3 + … + N-1 = N*(N-1)/2

However, because on each pass an average of only half of the maximum number of items are actually compared before the insertion point is found, we can divide by 2, which gives

N*(N-1)/4

The number of copies is approximately the same as the number of comparisons.

However, a copy isn’t as time-consuming as a swap, so for random data this algorithm runs twice as fast as the bubble sort and faster than the selection sort.

In any case, like the other sort routines in this chapter, the insertion sort runs in O(N2) time for random data.

For data that is already sorted or almost sorted, the insertion sort does much better.

When data is in order, the condition in the while loop is never true, so it becomes a simple statement in the outer loop, which executes N-1 times. In this case the algorithm runs in O(N) time. If the data is almost sorted, insertion sort runs in almost O(N) time, which makes it a simple and efficient way to order a file that is only slightly out of order.

However, for data arranged in inverse sorted order, every possible comparison and shift is carried out, so the insertion sort runs no faster than the bubble sort. You can check this using the reverse-sorted data option (toggled with New) in the InsertSort Workshop applet.

**Sorting Objects**

For simplicity we’ve applied the sorting algorithms we’ve looked at thus far to a primitive data type: long. However, sorting routines will more likely be applied to objects than primitive types. Accordingly, we show a Java program in Listing 3.4, objectSort.java, that sorts an array of Person objects (last seen in the classDataArray.java program in Chapter 2).

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**Java Code for Sorting Objects**

The algorithm used in our Java program is the insertion sort from the preceding section. The Person objects are sorted on lastName; this is the key field. The objectSort.java program is shown in Listing 3.4.

**LISTING 3.4**

The objectSort.java Program

// objectSort.java

// demonstrates sorting objects (uses insertion sort)

// to run this program: C>java ObjectSortApp

////////////////////////////////////////////////////////////////

class Person

{

private String lastName;

private String firstName;

private int age;

//-----------------------------------------------------------

public Person(String last, String first, int a)

{ // constructor

lastName = last;

firstName = first;

age = a;

}

//-----------------------------------------------------------

public void displayPerson()

{

System.out.print(“ Last name: “ + lastName);

System.out.print(“, First name: “ + firstName);

System.out.println(“, Age: “ + age);

}

//-----------------------------------------------------------

public String getLast() // get last name

{ return lastName; }

} // end class Person

////////////////////////////////////////////////////////////////

class ArrayInOb

{

private Person[] a; // ref to array a private int nElems; // number of data items

//--------------------------------------------------------------

public ArrayInOb(int max) // constructor

{

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**LISTING 3.4**

Continued

a = new Person[max]; // create the array nElems = 0; // no items yet

}

//--------------------------------------------------------------

// put person into array

public void insert(String last, String first, int age)

{

a[nElems] = new Person(last, first, age);

nElems++; // increment size

}

//--------------------------------------------------------------

public void display() // displays array contents

{

for(int j=0; j<nElems; j++) // for each element, a[j].displayPerson(); // display it

System.out.println(“”);

}

//--------------------------------------------------------------

public void insertionSort()

{

int in, out;

for(out=1; out<nElems; out++) // out is dividing line

{

Person temp = a[out]; // remove marked person in = out; // start shifting at out while(in>0 && // until smaller one found, a[in-1].getLast().compareTo(temp.getLast())>0)

{

a[in] = a[in-1]; // shift item to the right

--in; // go left one position

}

a[in] = temp; // insert marked item

} // end for

} // end insertionSort()

//--------------------------------------------------------------

} // end class ArrayInOb

////////////////////////////////////////////////////////////////

class ObjectSortApp

{

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**LISTING 3.4**

Continued

public static void main(String[] args)

{

int maxSize = 100; // array size

ArrayInOb arr; // reference to array arr = new ArrayInOb(maxSize); // create the array

arr.insert(“Evans”, “Patty”, 24);

arr.insert(“Smith”, “Doc”, 59);

arr.insert(“Smith”, “Lorraine”, 37);

arr.insert(“Smith”, “Paul”, 37);

arr.insert(“Yee”, “Tom”, 43);

arr.insert(“Hashimoto”, “Sato”, 21);

arr.insert(“Stimson”, “Henry”, 29);

arr.insert(“Velasquez”, “Jose”, 72);

arr.insert(“Vang”, “Minh”, 22);

arr.insert(“Creswell”, “Lucinda”, 18);

System.out.println(“Before sorting:”);

arr.display(); // display items

arr.insertionSort(); // insertion-sort them System.out.println(“After sorting:”);

arr.display(); // display them again

} // end main()

} // end class ObjectSortApp

////////////////////////////////////////////////////////////////

Here’s the output of this program:

Before sorting:

Last name: Evans, First name: Patty, Age: 24

Last name: Smith, First name: Doc, Age: 59

Last name: Smith, First name: Lorraine, Age: 37

Last name: Smith, First name: Paul, Age: 37

Last name: Yee, First name: Tom, Age: 43

Last name: Hashimoto, First name: Sato, Age: 21

Last name: Stimson, First name: Henry, Age: 29

Last name: Velasquez, First name: Jose, Age: 72

Last name: Vang, First name: Minh, Age: 22

Last name: Creswell, First name: Lucinda, Age: 18

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After sorting:

Last name: Creswell, First name: Lucinda, Age: 18

Last name: Evans, First name: Patty, Age: 24

Last name: Hashimoto, First name: Sato, Age: 21

Last name: Smith, First name: Doc, Age: 59

Last name: Smith, First name: Lorraine, Age: 37

Last name: Smith, First name: Paul, Age: 37

Last name: Stimson, First name: Henry, Age: 29

Last name: Vang, First name: Minh, Age: 22

Last name: Velasquez, First name: Jose, Age: 72

Last name: Yee, First name: Tom, Age: 43

**Lexicographical Comparisons**

The insertSort() method in objectSort.java is similar to that in insertSort.java, but it has been adapted to compare the lastName key values of records rather than the value of a primitive type.

We use the compareTo() method of the String class to perform the comparisons in the insertSort() method. Here’s the expression that uses it: a[in-1].getLast().compareTo(temp.getLast()) > 0

The compareTo() method returns different integer values depending on the lexicographical (that is, alphabetical) ordering of the String for which it’s invoked and the String passed to it as an argument, as shown in Table 3.1.

**TABLE 3.1**

Operation of the compareTo() Method

s2.compareTo(s1)

**Return Value**

s1 < s2

< 0

s1 equals s2

0

s1 > s2

> 0

For example, if s1 is “cat” and s2 is “dog”, the function will return a number less than 0. In the objectSort.java program, this method is used to compare the last name of a[in-1] with the last name of temp.

**Stability**

Sometimes it matters what happens to data items that have equal keys. For example, you may have employee data arranged alphabetically by last names. (That is, the last names were used as key values in the sort.) Now you want to sort the data by ZIP

code, but you want all the items with the same ZIP code to continue to be sorted by

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last names. You want the algorithm to sort only what needs to be sorted, and leave everything else in its original order. Some sorting algorithms retain this secondary ordering; they’re said to be

All the algorithms in this chapter are stable. For example, notice the output of the objectSort.java program (Listing 3.4). Three persons have the last name of Smith.

Initially, the order is Doc Smith, Lorraine Smith, and Paul Smith. After the sort, this ordering is preserved, despite the fact that the various Smith objects have been moved to new locations.

**Comparing the Simple Sorts**

There’s probably no point in using the bubble sort, unless you don’t have your algorithm book handy. The bubble sort is so simple that you can write it from memory.

Even so, it’s practical only if the amount of data is small. (For a discussion of what

“small” means, see Chapter 15, “When to Use What.”) The selection sort minimizes the number of swaps, but the number of comparisons is still high. This sort might be useful when the amount of data is small and swapping data items is very time-consuming compared with comparing them.

The insertion sort is the most versatile of the three and is the best bet in most situations, assuming the amount of data is small or the data is almost sorted. For larger amounts of data, quicksort is generally considered the fastest approach; we’ll examine quicksort in Chapter 7.

We’ve compared the sorting algorithms in terms of speed. Another consideration for any algorithm is how much memory space it needs. All three of the algorithms in this chapter carry out their sort

You can recompile the example programs, such as bubbleSort.java, to sort larger amounts of data. By timing them for larger sorts, you can get an idea of the differences between them and the time required to sort different amounts of data on your particular system.

**Summary**

• The sorting algorithms in this chapter all assume an array as a data storage structure.

• Sorting involves comparing the keys of data items in the array and moving the items (actually, references to the items) around until they’re in sorted order.

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• All the algorithms in this chapter execute in O(N2) time. Nevertheless, some can be substantially faster than others.

• An invariant is a condition that remains unchanged while an algorithm runs.

• The bubble sort is the least efficient, but the simplest, sort.

• The insertion sort is the most commonly used of the O(N2) sorts described in this chapter.

• A sort is stable if the order of elements with the same key is retained.

• None of the sorts in this chapter require more than a single temporary variable, in addition to the original array.

**Questions**

These questions are intended as a self-test for readers. Answers may be found in Appendix C.

**1. **Computer sorting algorithms are more limited than humans in that **a. **humans are better at inventing new algorithms.

**b. **computers can handle only a fixed amount of data.

**c. **humans know what to sort, whereas computers need to be told.

**d. **computers can compare only two things at a time.

**2. **The two basic operations in simple sorting are _________ items and _________

them (or sometimes _________ them).

**3. **True or False: The bubble sort always ends up comparing every item with every other item.

**4. **The bubble sort algorithm alternates between **a. **comparing and swapping.

**b. **moving and copying.

**c. **moving and comparing.

**d. **copying and comparing.

**5. **True or False: If there are N items, the bubble sort makes exactly N*N

comparisons.

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**6. **In the selection sort,

**a. **the largest keys accumulate on the left (low indices).

**b. **a minimum key is repeatedly discovered.

**c. **a number of items must be shifted to insert each item in its correctly sorted position.

**d. **the sorted items accumulate on the right.

**7. **True or False: If, in a particular sorting situation, swaps take much longer than comparisons, the selection sort is about twice as fast as the bubble sort.

**8. **A copy is ________ times as fast as a swap.

**9. **What is the invariant in the selection sort?

**10. **In the insertion sort, the “marked player” described in the text corresponds to which variable in the insertSort.java program?

**a. **in

**b. **out

**c. **temp

**d. **a[out]

**11. **In the insertion sort, “partially sorted” means that **a. **some items are already sorted, but they may need to be moved.

**b. **most items are in their final sorted positions, but a few still need to be sorted.

**c. **only some of the items are sorted.

**d. **group items are sorted among themselves, but items outside the group may need to be inserted in it.

**12. **Shifting a group of items left or right requires repeated __________.

**13. **In the insertion sort, after an item is inserted in the partially sorted group, it will

**a. **never be moved again.

**b. **never be shifted to the left.

**c. **often be moved out of this group.

**d. **find that its group is steadily shrinking.

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**14. **The invariant in the insertion sort is that ________.

**15. **Stability might refer to

**a. **items with secondary keys being excluded from a sort.

**b. **keeping cities sorted by increasing population within each state, in a sort by state.

**c. **keeping the same first names matched with the same last names.

**d. **items keeping the same order of primary keys without regard to secondary keys.

**Experiments**

Carrying out these experiments will help to provide insights into the topics covered in the chapter. No programming is involved.

**1. **In bubbleSort.java (Listing 3.1) rewrite main() so it creates a large array and fills that array with data. You can use the following code to generate random numbers:

for(int j=0; j<maxSize; j++) // fill array with

{ // random numbers long n = (long)( java.lang.Math.random()*(maxSize-1) ); arr.insert(n);

}

Try inserting 10,000 items. Display the data before and after the sort. You’ll see that scrolling the display takes a long time. Comment out the calls to display() so you can see how long the sort itself takes. The time will vary on different machines. Sorting 100,000 numbers will probably take less than 30 seconds.

Pick an array size that takes about this long and time it. Then use the same array size to time selectSort.java (Listing 3.2) and insertSort.java (Listing 3.3).

See how the speeds of these three sorts compare.

**2. **Devise some code to insert data in inversely sorted order (99,999, 99,998, 99,997, …) into bubbleSort.java. Use the same amount of data as in Experiment 1. See how fast the sort runs compared with the random data in Experiment 1. Repeat this experiment with selectSort.java and insertSort.java.

**3. **Write code to insert data in already-sorted order (0, 1, 2, …) into bubbleSort.java. See how fast the sort runs compared with Experiments 1 and 2. Repeat this experiment with selectSort.java and insertSort.java.

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**Programming Projects**

Writing programs that solve the Programming Projects helps to solidify your understanding of the material and demonstrates how the chapter’s concepts are applied.

(As noted in the Introduction, qualified instructors may obtain completed solutions to the Programming Projects on the publisher’s Web site.) **3.1 **In the bubbleSort.java program (Listing 3.1) and the BubbleSort Workshop applet, the in index always goes from left to right, finding the largest item and carrying it toward out on the right. Modify the bubbleSort() method so that it’s bidirectional. This means the in index will first carry the largest item from left to right as before, but when it reaches out, it will reverse and carry the smallest item from right to left. You’ll need two outer indexes, one on the right (the old out) and another on the left.

**3.2 **Add a method called median() to the ArrayIns class in the insertSort.java program (Listing 3.3). This method should return the median value in the array. (Recall that in a group of numbers half are larger than the median and half are smaller.) Do it the easy way.

**3.3 **To the insertSort.java program (Listing 3.3), add a method called noDups() that removes duplicates from a previously sorted array without disrupting the order.

(You can use the insertionSort() method to sort the data, or you can simply use main() to insert the data in sorted order.) One can imagine schemes in which all the items from the place where a duplicate was discovered to the end of the array would be shifted down one space every time a duplicate was discovered, but this would lead to slow O(N2) time, at least when there were a lot of duplicates. In your algorithm, make sure no item is moved more than once, no matter how many duplicates there are. This will give you an algorithm with O(N) time.

**3.4 **Another simple sort is the odd-even sort. The idea is to repeatedly make two passes through the array. On the first pass you look at all the pairs of items, a[j] and a[j+1], where j is odd (j = 1, 3, 5, …). If their key values are out of order, you swap them. On the second pass you do the same for all the even values (j = 2, 4, 6, …). You do these two passes repeatedly until the array is sorted. Replace the bubbleSort() method in bubbleSort.java (Listing 3.1) with an oddEvenSort() method. Make sure it works for varying amounts of data.

You’ll need to figure out how many times to do the two passes.

The odd-even sort is actually useful in a multiprocessing environment, where a separate processor can operate on each odd pair simultaneously and then on each even pair. Because the odd pairs are independent of each other, each pair can be checked—and swapped, if necessary—by a different processor. This makes for a very fast sort.

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**3.5 **Modify the insertionSort() method in insertSort.java (Listing 3.3) so it counts the number of copies and the number of comparisons it makes during a sort and displays the totals. To count comparisons, you’ll need to break up the double condition in the inner while loop. Use this program to measure the number of copies and comparisons for different amounts of inversely sorted data. Do the results verify O(N2) efficiency? Do the same for almost-sorted data (only a few items out of place). What can you deduce about the efficiency of this algorithm for almost-sorted data?

**3. **6 Here’s an interesting way to remove duplicates from an array. The insertion sort uses a loop-within-a-loop algorithm that compares every item in the array with every other item. If you want to remove duplicates, this is one way to start.

(See also Exercise 2.6 in Chapter 2.) Modify the insertionSort() method in the insertSort.java program so that it removes duplicates as it sorts. Here’s one approach: When a duplicate is found, write over one of the duplicated items with a key value less than any normally used (such as –1, if all the normal keys are positive). Then the normal insertion sort algorithm, treating this new key like any other item, will put it at index 0. From now on the algorithm can ignore this item. The next duplicate will go at index 1, and so on. When the sort is finished, all the removed dups (now represented by –1 values) will be found at the beginning of the array. The array can then be resized and shifted down so it starts at 0.

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**4**

**IN THIS CHAPTER**

• A Different Kind of Structure

Stacks and Queues

• Stacks

• Queues

• Priority Queues

In this chapter we’ll examine three data storage structures: the stack, the queue, and the priority queue. We’ll begin

• Parsing Arithmetic Expressions

by discussing how these structures differ from arrays; then we’ll examine each one in turn. In the last section, we’ll look at an operation in which the stack plays a significant role: parsing arithmetic expressions.

**A Different Kind of Structure**

There are significant differences between the data structures and algorithms we’ve seen in previous chapters and those we’ll look at now. We’ll discuss three of these differences before we examine the new structures in detail.

**Programmer’s Tools**

Arrays—the data storage structure we’ve been examining thus far—as well as many other structures we’ll encounter later in this book (linked lists, trees, and so on) are appropriate for the kind of data you might find in a database application. They’re typically used for personnel records, inventories, financial data, and so on—data that corresponds to real-world objects or activities. These structures facilitate access to data: They make it easy to insert, delete, and search for particular items.

The structures and algorithms we’ll examine in this chapter, on the other hand, are more often used as

programmer’s tools. They’re primarily conceptual aids rather than full-fledged data storage devices. Their lifetime is typically shorter than that of the database-type structures. They are created and used to carry out a particular task during the operation of a program; when the task is completed, they’re discarded.

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**Restricted Access**

In an array, any item can be accessed, either immediately—if its index number is known—or by searching through a sequence of cells until it’s found. In the data structures in this chapter, however, access is restricted: Only one item can be read or removed at a given time (unless you cheat).

The interface of these structures is designed to enforce this restricted access. Access to other items is (in theory) not allowed.

**More Abstract**

Stacks, queues, and priority queues are more abstract entities than arrays and many other data storage structures. They’re defined primarily by their interface: the permissible operations that can be carried out on them. The underlying mechanism used to implement them is typically not visible to their user.

The underlying mechanism for a stack, for example, can be an array, as shown in this chapter, or it can be a linked list. The underlying mechanism for a priority queue can be an array or a special kind of tree called a

**Stacks**

A stack allows access to only one data item: the last item inserted. If you remove this item, you can access the next-to-last item inserted, and so on. This capability is useful in many programming situations. In this section we’ll see how a stack can be used to check whether parentheses, braces, and brackets are balanced in a computer program source file. At the end of this chapter, we’ll see a stack playing a vital role in parsing (analyzing) arithmetic expressions such as 3*(4+5).

A stack is also a handy aid for algorithms applied to certain complex data structures.

In Chapter 8, “Binary Trees,” we’ll see it used to help traverse the nodes of a tree. In Chapter 13, “Graphs,” we’ll apply it to searching the vertices of a graph (a technique that can be used to find your way out of a maze).

Most microprocessors use a stack-based architecture. When a method is called, its return address and arguments are pushed onto a stack, and when it returns, they’re popped off. The stack operations are built into the microprocessor.

Some older pocket calculators used a stack-based architecture. Instead of entering arithmetic expressions using parentheses, you pushed intermediate results onto a stack. We’ll learn more about this approach when we discuss parsing arithmetic expressions in the last section in this chapter.

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**The Postal Analogy**

To understand the idea of a stack, consider an analogy provided by the U.S. Postal Service. Many people, when they get their mail, toss it onto a stack on the hall table or into an “in” basket at work. Then, when they have a spare moment, they process the accumulated mail from the top down. First, they open the letter on the top of the stack and take appropriate action—paying the bill, throwing it away, or whatever. After the first letter has been disposed of, they examine the next letter down, which is now the top of the stack, and deal with that. Eventually, they work their way down to the letter on the bottom of the stack (which is now the top). Figure 4.1

shows a stack of mail.

This letter

processed first

Newly arrived

letters placed

on top of

stack

**FIGURE 4.1**

A stack of letters.

This “do the top one first” approach works all right as long as you can easily process all the mail in a reasonable time. If you can’t, there’s the danger that letters on the bottom of the stack won’t be examined for months, and the bills they contain will become overdue.

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Of course, many people don’t rigorously follow this top-to-bottom approach. They may, for example, take the mail off the bottom of the stack, so as to process the oldest letter first. Or they might shuffle through the mail before they begin processing it and put higher-priority letters on top. In these cases, their mail system is no longer a stack in the computer-science sense of the word. If they take letters off the bottom, it’s a queue; and if they prioritize it, it’s a priority queue. We’ll look at these possibilities later.

Another stack analogy is the tasks you perform during a typical workday. You’re busy on a long-term project (A), but you’re interrupted by a coworker asking you for temporary help with another project (B). While you’re working on B, someone in accounting stops by for a meeting about travel expenses (C), and during this meeting you get an emergency call from someone in sales and spend a few minutes trou-bleshooting a bulky product (D). When you’re done with call D, you resume meeting C; when you’re done with C, you resume project B, and when you’re done with B, you can (finally!) get back to project A. Lower-priority projects are “stacked up”

waiting for you to return to them.

Placing a data item on the top of the stack is called

**The Stack Workshop Applet**

Let’s use the Stack Workshop applet to get an idea how stacks work. When you start up this applet, you’ll see four buttons: New, Push, Pop, and Peek, as shown in Figure 4.2.

**FIGURE 4.2**

The Stack Workshop applet.

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The Stack Workshop applet is based on an array, so you’ll see an array of data items.

Although it’s based on an array, a stack restricts access, so you can’t access elements using an index. In fact, the concept of a stack and the underlying data structure used to implement it are quite separate. As we noted earlier, stacks can also be implemented by other kinds of storage structures, such as linked lists.

**The New Button**

The stack in the Workshop applet starts off with four data items already inserted. If you want to start with an empty stack, the New button creates a new stack with no items. The next three buttons carry out the significant stack operations.

**The Push Button**

To insert a data item on the stack, use the button labeled Push. After the first press of this button, you’ll be prompted to enter the key value of the item to be pushed.

After you type the value into the text field, a few more presses will insert the item on the top of the stack.

A red arrow always points to the top of the stack—that is, the last item inserted.

Notice how, during the insertion process, one step (button press) increments (moves up) the Top arrow, and the next step actually inserts the data item into the cell. If you reversed the order, you would overwrite the existing item at Top. When you’re writing the code to implement a stack, it’s important to keep in mind the order in which these two steps are executed.

If the stack is full and you try to push another item, you’ll get the Can’t insert: stack is full message. (Theoretically, an ADT stack doesn’t become full, but the array implementing it does.)

**The Pop Button**

To remove a data item from the top of the stack, use the Pop button. The value popped appears in the Number text field; this corresponds to a pop() routine returning a value.

Again, notice the two steps involved: First, the item is removed from the cell pointed to by Top; then Top is decremented to point to the highest occupied cell. This is the reverse of the sequence used in the push operation.

The pop operation shows an item actually being removed from the array and the cell color becoming gray to show the item has been removed. This is a bit misleading, in that deleted items actually remain in the array until written over by new data.

However, they cannot be accessed after the Top marker drops below their position, so conceptually they are gone, as the applet shows.

After you’ve popped the last item off the stack, the Top arrow points to –1, below the lowest cell. This position indicates that the stack is empty. If the stack is empty and you try to pop an item, you’ll get the Can’t pop: stack is empty message.

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**The Peek Button**

Push and pop are the two primary stack operations. However, it’s sometimes useful to be able to read the value from the top of the stack without removing it. The peek operation does this. By pushing the Peek button a few times, you’ll see the value of the item at Top copied to the Number text field, but the item is not removed from the stack, which remains unchanged.

Notice that you can peek only at the top item. By design, all the other items are invisible to the stack user.

**Stack Size**

Stacks are typically small, temporary data structures, which is why we’ve shown a stack of only 10 cells. Of course, stacks in real programs may need a bit more room than this, but it’s surprising how small a stack needs to be. A very long arithmetic expression, for example, can be parsed with a stack of only a dozen or so cells.

**Java Code for a Stack**

Let’s examine a program, stack.java, that implements a stack using a class called StackX. Listing 4.1 contains this class and a short main() routine to exercise it.

**LISTING 4.1**

The stack.java Program

// stack.java

// demonstrates stacks

// to run this program: C>java StackApp

////////////////////////////////////////////////////////////////

class StackX

{

private int maxSize; // size of stack array

private long[] stackArray;

private int top; // top of stack

//--------------------------------------------------------------

public StackX(int s) // constructor

{

maxSize = s; // set array size

stackArray = new long[maxSize]; // create array

top = -1; // no items yet

}

//--------------------------------------------------------------

public void push(long j) // put item on top of stack

{

stackArray[++top] = j; // increment top, insert item

}

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**LISTING 4.1**

Continued

//--------------------------------------------------------------

public long pop() // take item from top of stack

{

return stackArray[top--]; // access item, decrement top

}

//--------------------------------------------------------------

public long peek() // peek at top of stack

{

return stackArray[top];

}

//--------------------------------------------------------------

public boolean isEmpty() // true if stack is empty

{

return (top == -1);

}

//--------------------------------------------------------------

public boolean isFull() // true if stack is full

{

return (top == maxSize-1);

}

//--------------------------------------------------------------

} // end class StackX

////////////////////////////////////////////////////////////////

class StackApp

{

public static void main(String[] args)

{

StackX theStack = new StackX(10); // make new stack theStack.push(20); // push items onto stack theStack.push(40);

theStack.push(60);

theStack.push(80);

while( !theStack.isEmpty() ) // until it’s empty,

{ // delete item from stack long value = theStack.pop();

System.out.print(value); // display it

System.out.print(“ “);

} // end while

System.out.println(“”);

} // end main()

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**LISTING 4.1**

Continued

} // end class StackApp

////////////////////////////////////////////////////////////////

The main() method in the StackApp class creates a stack that can hold 10 items, pushes 4 items onto the stack, and then displays all the items by popping them off the stack until it’s empty. Here’s the output:

80 60 40 20

Notice how the order of the data is reversed. Because the last item pushed is the first one popped, the 80 appears first in the output.

This version of the StackX class holds data elements of type long. As noted in Chapter 3, “Simple Sorting,” you can change this to any other type, including object types.

StackX **Class Methods**

The constructor creates a new stack of a size specified in its argument. The fields of the stack are made up of a variable to hold its maximum size (the size of the array), the array itself, and a variable top, which stores the index of the item on the top of the stack. (Note that we need to specify a stack size only because the stack is implemented using an array. If it had been implemented using a linked list, for example, the size specification would be unnecessary.)

The push() method increments top so it points to the space just above the previous top and stores a data item there. Notice again that top is incremented before the item is inserted.

The pop() method returns the value at top and then decrements top. This effectively removes the item from the stack; it’s inaccessible, although the value remains in the array (until another item is pushed into the cell).

The peek() method simply returns the value at top, without changing the stack.

The isEmpty() and isFull() methods return true if the stack is empty or full, respectively. The top variable is at –1 if the stack is empty and maxSize-1 if the stack is full.

Figure 4.3 shows how the stack class methods work.

**Error Handling**

There are different philosophies about how to handle stack errors. What happens if you try to push an item onto a stack that’s already full or pop an item from a stack that’s empty?

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123

49

Top

49

Top

27

27

14

14

3

3

92

92

64

64

New item pushed on stack

49

49

Top

27

27

Top

27

14

14

Top

14

3

3

3

92

92

92

64

64

64

Two items popped from stack

**FIGURE 4.3**

Operation of the StackX class methods.

We’ve left the responsibility for handling such errors up to the class user. The user should always check to be sure the stack is not full before inserting an item: if( !theStack.isFull() )

insert(item);

else

System.out.print(“Can’t insert, stack is full”);

In the interest of simplicity, we’ve left this code out of the main() routine (and anyway, in this simple program, we know the stack isn’t full because it has just been initialized). We do include the check for an empty stack when main() calls pop().

Many stack classes check for these errors internally, in the push() and pop() methods.

This is the preferred approach. In Java, a good solution for a stack class that discovers such errors is to throw an exception, which can then be caught and processed by the class user.

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**Stack Example 1: Reversing a Word**

For our first example of using a stack, we’ll examine a very simple task: reversing a word. When you run the program, it asks you to type in a word. When you press Enter, it displays the word with the letters in reverse order.

A stack is used to reverse the letters. First, the characters are extracted one by one from the input string and pushed onto the stack. Then they’re popped off the stack and displayed. Because of its Last-In-First-Out characteristic, the stack reverses the order of the characters. Listing 4.2 shows the code for the reverse.java program.

**LISTING 4.2**

The reverse.java Program

// reverse.java

// stack used to reverse a string

// to run this program: C>java ReverseApp

import java.io.*; // for I/O

////////////////////////////////////////////////////////////////

class StackX

{

private int maxSize;

private char[] stackArray;

private int top;

//--------------------------------------------------------------

public StackX(int max) // constructor

{

maxSize = max;

stackArray = new char[maxSize];

top = -1;

}

//--------------------------------------------------------------

public void push(char j) // put item on top of stack

{

stackArray[++top] = j;

}

//--------------------------------------------------------------

public char pop() // take item from top of stack

{

return stackArray[top--];

}

//--------------------------------------------------------------

public char peek() // peek at top of stack

{

return stackArray[top];

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**LISTING 4.2**

Continued

}

//--------------------------------------------------------------

public boolean isEmpty() // true if stack is empty

{

return (top == -1);

}

//--------------------------------------------------------------

} // end class StackX

////////////////////////////////////////////////////////////////

class Reverser

{

private String input; // input string private String output; // output string

//--------------------------------------------------------------

public Reverser(String in) // constructor

{ input = in; }

//--------------------------------------------------------------

public String doRev() // reverse the string

{

int stackSize = input.length(); // get max stack size StackX theStack = new StackX(stackSize); // make stack for(int j=0; j<input.length(); j++)

{

char ch = input.charAt(j); // get a char from input theStack.push(ch); // push it

}

output = “”;

while( !theStack.isEmpty() )

{

char ch = theStack.pop(); // pop a char,

output = output + ch; // append to output

}

return output;

} // end doRev()

//--------------------------------------------------------------

} // end class Reverser

////////////////////////////////////////////////////////////////

class ReverseApp

{

public static void main(String[] args) throws IOException

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**LISTING 4.2**

Continued

{

String input, output;

while(true)

{

System.out.print(“Enter a string: “);

System.out.flush();

input = getString(); // read a string from kbd if( input.equals(“”) ) // quit if [Enter]

break;

// make a Reverser

Reverser theReverser = new Reverser(input);

output = theReverser.doRev(); // use it

System.out.println(“Reversed: “ + output);

} // end while

} // end main()

//--------------------------------------------------------------

public static String getString() throws IOException

{

InputStreamReader isr = new InputStreamReader(System.in); BufferedReader br = new BufferedReader(isr);

String s = br.readLine();

return s;

}

//--------------------------------------------------------------

} // end class ReverseApp

////////////////////////////////////////////////////////////////

We’ve created a class Reverser to handle the reversing of the input string. Its key component is the method doRev(), which carries out the reversal, using a stack. The stack is created within doRev(), which sizes the stack according to the length of the input string.

In main() we get a string from the user, create a Reverser object with this string as an argument to the constructor, call this object’s doRev() method, and display the return value, which is the reversed string. Here’s some sample interaction with the program: Enter a string: part

Reversed: trap

Enter a string:

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**Stack Example 2: Delimiter Matching**

One common use for stacks is to parse certain kinds of text strings. Typically, the strings are lines of code in a computer language, and the programs parsing them are compilers.

To give the flavor of what’s involved, we’ll show a program that checks the delimiters in a line of text typed by the user. This text doesn’t need to be a line of real Java code (although it could be), but it should use delimiters the same way Java does. The delimiters are the braces { and }, brackets [ and ], and parentheses ( and ). Each opening or left delimiter should be matched by a closing or right delimiter; that is, every { should be followed by a matching } and so on. Also, opening delimiters that occur later in the string should be closed before those occurring earlier. Here are some examples:

c[d] // correct

a{b[c]d}e // correct

a{b(c]d}e // not correct; ] doesn’t match (

a[b{c}d]e} // not correct; nothing matches final }

a{b(c) // not correct; nothing matches opening {

**Opening Delimiters on the Stack**

This delimiter-matching program works by reading characters from the string one at a time and placing opening delimiters when it finds them, on a stack. When it reads a closing delimiter from the input, it pops the opening delimiter from the top of the stack and attempts to match it with the closing delimiter. If they’re not the same type (there’s an opening brace but a closing parenthesis, for example), an error occurs. Also, if there is no opening delimiter on the stack to match a closing one, or if a delimiter has not been matched, an error occurs. A delimiter that hasn’t been matched is discovered because it remains on the stack after all the characters in the string have been read.

Let’s see what happens on the stack for a typical correct string: a{b(c[d]e)f}

Table 4.1 shows how the stack looks as each character is read from this string. The entries in the second column show the stack contents, reading from the bottom of the stack on the left to the top on the right.

As the string is read, each opening delimiter is placed on the stack. Each closing delimiter read from the input is matched with the opening delimiter popped from the top of the stack. If they form a pair, all is well. Non-delimiter characters are not inserted on the stack; they’re ignored.

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**TABLE 4.1**

Stack Contents in Delimiter Matching

**Character Read**

**Stack Contents**

a

{

{

b

{

(

{(

c

{(

[

{([

d

{([

]

{(

e

{(

)

{

f

{

}

This approach works because pairs of delimiters that are opened last should be closed first. This matches the Last-In-First-Out property of the stack.

**Java Code for **brackets.java

The code for the parsing program, brackets.java, is shown in Listing 4.3. We’ve placed check(), the method that does the parsing, in a class called BracketChecker.

**LISTING 4.3**

The brackets.java Program

// brackets.java

// stacks used to check matching brackets

// to run this program: C>java BracketsApp

import java.io.*; // for I/O

////////////////////////////////////////////////////////////////

class StackX

{

private int maxSize;

private char[] stackArray;

private int top;

//--------------------------------------------------------------

public StackX(int s) // constructor

{

maxSize = s;

stackArray = new char[maxSize];

top = -1;

}

//--------------------------------------------------------------

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**LISTING 4.3**

Continued

public void push(char j) // put item on top of stack

{

stackArray[++top] = j;

}

//--------------------------------------------------------------

public char pop() // take item from top of stack

{

return stackArray[top--];

}

//--------------------------------------------------------------

public char peek() // peek at top of stack

{

return stackArray[top];

}

//--------------------------------------------------------------

public boolean isEmpty() // true if stack is empty

{

return (top == -1);

}

//--------------------------------------------------------------

} // end class StackX

////////////////////////////////////////////////////////////////

class BracketChecker

{

private String input; // input string

//--------------------------------------------------------------

public BracketChecker(String in) // constructor

{ input = in; }

//--------------------------------------------------------------

public void check()

{

int stackSize = input.length(); // get max stack size StackX theStack = new StackX(stackSize); // make stack for(int j=0; j<input.length(); j++) // get chars in turn

{

char ch = input.charAt(j); // get char

switch(ch)

{

case ‘{‘: // opening symbols

case ‘[‘:

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**LISTING 4.3**

Continued

case ‘(‘:

theStack.push(ch); // push them

break;

case ‘}’: // closing symbols

case ‘]’:

case ‘)’:

if( !theStack.isEmpty() ) // if stack not empty,

{

char chx = theStack.pop(); // pop and check

if( (ch==’}’ && chx!=’{‘) ||

(ch==’]’ && chx!=’[‘) ||

(ch==’)’ && chx!=’(‘) )

System.out.println(“Error: “+ch+” at “+j);

}

else // prematurely empty

System.out.println(“Error: “+ch+” at “+j);

break;

default: // no action on other characters

break;

} // end switch

} // end for

// at this point, all characters have been processed if( !theStack.isEmpty() )

System.out.println(“Error: missing right delimiter”);

} // end check()

//--------------------------------------------------------------

} // end class BracketChecker

////////////////////////////////////////////////////////////////

class BracketsApp

{

public static void main(String[] args) throws IOException

{

String input;

while(true)

{

System.out.print(

“Enter string containing delimiters: “);

System.out.flush();

input = getString(); // read a string from kbd

if( input.equals(“”) ) // quit if [Enter]

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**LISTING 4.3**

Continued

break;

// make a BracketChecker

BracketChecker theChecker = new BracketChecker(input); theChecker.check(); // check brackets

} // end while

} // end main()

//--------------------------------------------------------------

public static String getString() throws IOException

{

InputStreamReader isr = new InputStreamReader(System.in); BufferedReader br = new BufferedReader(isr);

String s = br.readLine();

return s;

}

//--------------------------------------------------------------

} // end class BracketsApp

////////////////////////////////////////////////////////////////

The check() routine makes use of the StackX class from the reverse.java program (Listing 4.2). Notice how easy it is to reuse this class. All the code you need is in one place. This is one of the payoffs for object-oriented programming.

The main() routine in the BracketsApp class repeatedly reads a line of text from the user, creates a BracketChecker object with this text string as an argument, and then calls the check() method for this BracketChecker object. If it finds any errors, the check() method displays them; otherwise, the syntax of the delimiters is correct.

If it can, the check() method reports the character number where it discovered the error (starting at 0 on the left) and the incorrect character it found there. For example, for the input string

a{b(c]d}e

the output from check() will be

Error: ] at 5

**The Stack as a Conceptual Aid**

Notice how convenient the stack is in the brackets.java program. You could have set up an array to do what the stack does, but you would have had to worry about keeping track of an index to the most recently added character, as well as other bookkeeping tasks. The stack is conceptually easier to use. By providing limited access to its contents, using the push() and pop() methods, the stack has made your

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program easier to understand and less error prone. (As carpenters will tell you, it’s safer to use the right tool for the job.)

**Efficiency of Stacks**

Items can be both pushed and popped from the stack implemented in the StackX

class in constant O(1) time. That is, the time is not dependent on how many items are in the stack and is therefore very quick. No comparisons or moves are necessary.

**Queues**

The word

means to get in line. In computer science a queue is a data structure that is somewhat like a stack, except that in a queue the first item inserted is the first to be removed (First-In-First-Out, FIFO), while in a stack, as we’ve seen, the last item inserted is the first to be removed (LIFO). A queue works like the line at the movies: The first person to join the rear of the line is the first person to reach the front of the line and buy a ticket. The last person to line up is the last person to buy a ticket (or—if the show is sold out—to fail to buy a ticket). Figure 4.4 shows how such a queue looks.

People join the

queue at the rear

People leave the

queue at the front

**FIGURE 4.4**

A queue of people.

Queues are used as a programmer’s tool as stacks are. We’ll see an example where a queue helps search a graph in Chapter 13. They’re also used to model real-world situations such as people waiting in line at a bank, airplanes waiting to take off, or data packets waiting to be transmitted over the Internet.

There are various queues quietly doing their job in your computer’s (or the network’s) operating system. There’s a printer queue where print jobs wait for the

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printer to be available. A queue also stores keystroke data as you type at the keyboard. This way, if you’re using a word processor but the computer is briefly doing something else when you hit a key, the keystroke won’t be lost; it waits in the queue until the word processor has time to read it. Using a queue guarantees the keystrokes stay in order until they can be processed.

**The Queue Workshop Applet**

Let’s use the Queue Workshop applet to get an idea how queues work. When you start up the applet, you’ll see a queue with four items preinstalled, as shown in Figure 4.5.

**FIGURE 4.5**

The Queue Workshop applet.

This applet demonstrates a queue based on an array. This is a common approach, although linked lists are also commonly used to implement queues.

The two basic queue operations are

The terms for insertion and removal in a stack are fairly standard; everyone says

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**The Insert Button**

By repeatedly pressing the Ins button in the Queue Workshop applet, you can insert a new item. After the first press, you’re prompted to enter a key value for a new item into the Number text field; this should be a number from 0 to 999. Subsequent presses will insert an item with this key at the rear of the queue and increment the Rear arrow so it points to the new item.

**The Remove Button**

Similarly, you can remove the item at the front of the queue using the Rem button.

The item is removed, the item’s value is stored in the Number field (corresponding to the remove() method returning a value), and the Front arrow is incremented. In the applet, the cell that held the deleted item is grayed to show it’s gone. In a normal implementation, it would remain in memory but would not be accessible because Front had moved past it. The insert and remove operations are shown in Figure 4.6.

6

Rear

6

Rear

80

80

12

12

94

96

26

26

Front

59

Front

59

New item inserted at rear of queue

6

6

Rear

6

Rear

Rear

80

80

80

12

12

12

94

94

Front

94

Front

26

26

59

Front

26

59

Two items removed from front of queue

**FIGURE 4.6**

Operation of the Queue class methods.

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Unlike the situation in a stack, the items in a queue don’t always extend all the way down to index 0 in the array. After some items are removed, Front will point at a cell with a higher index, as shown in Figure 4.7.

In Figure 4.7, notice that Front lies below Rear in the array; that is, Front has a lower index. As we’ll see in a moment, this isn’t always true.

**The Peek Button**

We show one other queue operation, peek. Peek finds the value of the item at the front of the queue without removing the item. (Like insert and remove, peek, when applied to a queue, is also called by a variety of other names.) If you press the Peek button, you’ll see the value at Front transferred to the Number field. The queue is unchanged. This peek() method returns the value at the front of the queue. Some queue implementations have a rearPeek() and a frontPeek() method, but usually you want to know what you’re about to remove, not what you just inserted.

MaxSize-1

9

Empty cells

8

7

79

Rear

6

32

5

6

4

80

3

12

Front

2

1

Empty cells

0

**FIGURE 4.7**

A queue with some items removed.

**The New Button**

If you want to start with an empty queue, you can use the New button to create one.

**Empty and Full**

If you try to remove an item when there are no more items in the queue, you’ll get the Can’t remove, queue is empty error message. If you try to insert an item when all the cells are already occupied, you’ll get the Can’t insert, queue is full message.

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**A Circular Queue**

When you insert a new item in the queue in the Queue Workshop applet, the Front arrow moves upward, toward higher numbers in the array. When you remove an item, Rear also moves upward. Try these operations with the Workshop applet to convince yourself it’s true. You may find the arrangement counter-intuitive, because the people in a line at the movies all move forward, toward the front, when a person leaves the line. We could move all the items in a queue whenever we deleted one, but that wouldn’t be very efficient. Instead, we keep all the items in the same place and move the front and rear of the queue.

The trouble with this arrangement is that pretty soon the rear of the queue is at the end of the array (the highest index). Even if there are empty cells at the beginning of the array, because you’ve removed them with Rem, you still can’t insert a new item because Rear can’t go any further. Or can it? This situation is shown in Figure 4.8.

MaxSize-1

9

44

Rear

8

21

7

79

63

6

32

5

6

New item:

4

80

Where can

it go?

3

12

Front

2

1

0

**FIGURE 4.8**

Rear arrow at the end of the array.

**Wrapping Around**

To avoid the problem of not being able to insert more items into the queue even when it’s not full, the Front and Rear arrows

You can see how wraparound works with the Workshop applet. Insert enough items to bring the Rear arrow to the top of the array (index 9). Remove some items from

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137

the front of the array. Now insert another item. You’ll see the Rear arrow wrap around from index 9 to index 0; the new item will be inserted there. This situation is shown in Figure 4.9.

Insert a few more items. The Rear arrow moves upward as you’d expect. Notice that after Rear has wrapped around, it’s now below Front, the reverse of the original arrangement. You can call this a

MaxSize-1

9

44

8

21

7

79

6

32

5

6

4

80

3

12

Front

2

1

0

63

Rear

**FIGURE 4.9**

The Rear arrow wraps around.

Delete enough items so that the Front arrow also wraps around. Now you’re back to the original arrangement, with Front below Rear. The items are in a single

**Java Code for a Queue**

The queue.java program features a Queue class with insert(), remove(), peek(), isFull(), isEmpty(), and size() methods.

The main() program creates a queue of five cells, inserts four items, removes three items, and inserts four more. The sixth insertion invokes the wraparound feature. All the items are then removed and displayed. The output looks like this: 40 50 60 70 80

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Listing 4.4 shows the queue.java program.

**LISTING 4.4**

The queue.java Program

// queue.java

// demonstrates queue

// to run this program: C>java QueueApp

////////////////////////////////////////////////////////////////

class Queue

{

private int maxSize;

private long[] queArray;

private int front;

private int rear;

private int nItems;

//--------------------------------------------------------------

public Queue(int s) // constructor

{

maxSize = s;

queArray = new long[maxSize];

front = 0;

rear = -1;

nItems = 0;

}

//--------------------------------------------------------------

public void insert(long j) // put item at rear of queue

{

if(rear == maxSize-1) // deal with wraparound rear = -1;

queArray[++rear] = j; // increment rear and insert nItems++; // one more item

}

//--------------------------------------------------------------

public long remove() // take item from front of queue

{

long temp = queArray[front++]; // get value and incr front if(front == maxSize) // deal with wraparound front = 0;

nItems--; // one less item

return temp;

}

//--------------------------------------------------------------

public long peekFront() // peek at front of queue

{

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**LISTING 4.4**

Continued

return queArray[front];

}

//--------------------------------------------------------------

public boolean isEmpty() // true if queue is empty

{

return (nItems==0);

}

//--------------------------------------------------------------

public boolean isFull() // true if queue is full

{

return (nItems==maxSize);

}

//--------------------------------------------------------------

public int size() // number of items in queue

{

return nItems;

}

//--------------------------------------------------------------

} // end class Queue

////////////////////////////////////////////////////////////////

class QueueApp

{

public static void main(String[] args)

{

Queue theQueue = new Queue(5); // queue holds 5 items theQueue.insert(10); // insert 4 items

theQueue.insert(20);

theQueue.insert(30);

theQueue.insert(40);

theQueue.remove(); // remove 3 items

theQueue.remove(); // (10, 20, 30)

theQueue.remove();

theQueue.insert(50); // insert 4 more items theQueue.insert(60); // (wraps around) theQueue.insert(70);

theQueue.insert(80);

while( !theQueue.isEmpty() ) // remove and display

{ // all items

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**LISTING 4.4**

Continued

long n = theQueue.remove(); // (40, 50, 60, 70, 80) System.out.print(n);

System.out.print(“ “);

}

System.out.println(“”);

} // end main()

} // end class QueueApp

We’ve chosen an approach in which Queue class fields include not only front and rear, but also the number of items currently in the queue: nItems. Some queue implementations don’t use this field; we’ll show this alternative later.

**The **insert() **Method**

The insert() method assumes that the queue is not full. We don’t show it in main(), but normally you should call insert() only after calling isFull() and getting a return value of false. (It’s usually preferable to place the check for fullness in the insert() routine and cause an exception to be thrown if an attempt was made to insert into a full queue.)

Normally, insertion involves incrementing rear and inserting at the cell rear now points to. However, if rear is at the top of the array, at maxSize-1, then it must wrap around to the bottom of the array before the insertion takes place. This is done by setting rear to –1, so when the increment occurs, rear will become 0, the bottom of the array. Finally, nItems is incremented.

**The **remove() **Method**

The remove() method assumes that the queue is not empty. You should call isEmpty() to ensure this is true before calling remove(), or build this error-checking into remove().

Removal always starts by obtaining the value at front and then incrementing front.

However, if this puts front beyond the end of the array, it must then be wrapped around to 0. The return value is stored temporarily while this possibility is checked.

Finally, nItems is decremented.

**The **peek() **Method**

The peek() method is straightforward: It returns the value at front. Some implementations allow peeking at the rear of the array as well; such routines are called something like peekFront() and peekRear() or just front() and rear().

**The **isEmpty()**, **isFull()**, and **size() **Methods** The isEmpty(), isFull(), and size() methods all rely on the nItems field, respectively checking if it’s 0, if it’s maxSize, or returning its value.

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**Implementation Without an Item Count**

The inclusion of the field nItems in the Queue class imposes a slight overhead on the insert() and remove() methods in that they must respectively increment and decrement this variable. This may not seem like an excessive penalty, but if you’re dealing with huge numbers of insertions and deletions, it might influence performance.

Accordingly, some implementations of queues do without an item count and rely on the front and rear fields to figure out whether the queue is empty or full and how many items are in it. When this is done, the isEmpty(), isFull(), and size() routines become surprisingly complicated because the sequence of items may be either broken or contiguous, as we’ve seen.

Also, a strange problem arises. The front and rear pointers assume certain positions when the queue is full, but they can assume these exact same positions when the queue is empty. The queue can then appear to be full and empty at the same time.

This problem can be solved by making the array one cell larger than the maximum number of items that will be placed in it. Listing 4.5 shows a Queue class that implements this no-count approach. This class uses the no-count implementation.

**LISTING 4.5**

The Queue Class Without nItems

class Queue

{

private int maxSize;

private long[] queArray;

private int front;

private int rear;

//--------------------------------------------------------------

public Queue(int s) // constructor

{

maxSize = s+1; // array is 1 cell larger queArray = new long[maxSize]; // than requested

front = 0;

rear = -1;

}

//--------------------------------------------------------------

public void insert(long j) // put item at rear of queue

{

if(rear == maxSize-1)

rear = -1;

queArray[++rear] = j;

}

//--------------------------------------------------------------

public long remove() // take item from front of queue

{

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**LISTING 4.5**

Continued

long temp = queArray[front++];

if(front == maxSize)

front = 0;

return temp;

}

//--------------------------------------------------------------

public long peek() // peek at front of queue

{

return queArray[front];

}

//--------------------------------------------------------------

public boolean isEmpty() // true if queue is empty

{

return ( rear+1==front || (front+maxSize-1==rear) );

}

//--------------------------------------------------------------

public boolean isFull() // true if queue is full

{

return ( rear+2==front || (front+maxSize-2==rear) );

}

//--------------------------------------------------------------

public int size() // (assumes queue not empty)

{

if(rear >= front) // contiguous sequence return rear-front+1;

else // broken sequence

return (maxSize-front) + (rear+1);

}

//--------------------------------------------------------------

} // end class Queue

Notice the complexity of the isFull(), isEmpty(), and size() methods. This no-count approach is seldom needed in practice, so we’ll refrain from discussing it in detail.

**Efficiency of Queues**

As with a stack, items can be inserted and removed from a queue in O(1) time.

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143

**Deques**

A

If you restrict yourself to insertLeft() and removeLeft() (or their equivalents on the right), the deque acts like a stack. If you restrict yourself to insertLeft() and removeRight() (or the opposite pair), it acts like a queue.

A deque provides a more versatile data structure than either a stack or a queue and is sometimes used in container class libraries to serve both purposes. However, it’s not used as often as stacks and queues, so we won’t explore it further here.

**Priority Queues**

A priority queue is a more specialized data structure than a stack or a queue.

However, it’s a useful tool in a surprising number of situations. Like an ordinary queue, a priority queue has a front and a rear, and items are removed from the front.

However, in a priority queue, items are ordered by key value so that the item with the lowest key (or in some implementations the highest key) is always at the front.

Items are inserted in the proper position to maintain the order.

Here’s how the mail sorting analogy applies to a priority queue. Every time the postman hands you a letter, you insert it into your pile of pending letters according to its priority. If it must be answered immediately (the phone company is about to disconnect your modem line), it goes on top, whereas if it can wait for a leisurely answer (a letter from your Aunt Mabel), it goes on the bottom. Letters with intermediate priorities are placed in the middle; the higher the priority, the higher their position in the pile. The top of the pile of letters corresponds to the front of the priority queue.

When you have time to answer your mail, you start by taking the letter off the top (the front of the queue), thus ensuring that the most important letters are answered first. This situation is shown in Figure 4.10.

Like stacks and queues, priority queues are often used as programmer’s tools. We’ll see one used in finding something called a minimum spanning tree for a graph, in Chapter 14, “Weighted Graphs.”

Also, like ordinary queues, priority queues are used in various ways in certain computer systems. In a preemptive multitasking operating system, for example, programs may be placed in a priority queue so the highest-priority program is the next one to receive a time-slice that allows it to execute.

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Letter on top

is always

processed

first

More urgent letters are

inserted higher

Less urgent letters are

inserted lower

**FIGURE 4.10 **

In many situations you want access to the item with the lowest key value (which might represent the cheapest or shortest way to do something). Thus, the item with the smallest key has the highest priority. Somewhat arbitrarily, we’ll assume that’s the case in this discussion, although there are other situations in which the highest key has the highest priority.

Besides providing quick access to the item with the smallest key, you also want a priority queue to provide fairly quick insertion. For this reason, priority queues are, as we noted earlier, often implemented with a data structure called a heap. We’ll look at heaps in Chapter 12, “Heaps.” In this chapter, we’ll show a priority queue implemented by a simple array. This implementation suffers from slow insertion, but it’s simpler and is appropriate when the number of items isn’t high or insertion speed isn’t critical.

**The PriorityQ Workshop Applet**

The PriorityQ Workshop applet implements a priority queue with an array, in which the items are kept in sorted order. It’s an

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The minimum-key item is always at the top (highest index) in the array, and the largest item is always at index 0. Figure 4.11 shows the arrangement when the applet is started. Initially, there are five items in the queue.

**FIGURE 4.11 **

**The Insert Button**

Try inserting an item. You’ll be prompted to type the new item’s key value into the Number field. Choose a number that will be inserted somewhere in the middle of the values already in the queue. For example, in Figure 4.11 you might choose 300.

Then, as you repeatedly press Ins, you’ll see that the items with smaller keys are shifted up to make room. A black arrow shows which item is being shifted. When the appropriate position is found, the new item is inserted into the newly created space.

Notice that there’s no wraparound in this implementation of the priority queue.

Insertion is slow of necessity because the proper in-order position must be found, but deletion is fast. A wraparound implementation wouldn’t improve the situation.

Note too that the Rear arrow never moves; it always points to index 0 at the bottom of the array.

**The Delete Button**

The item to be removed is always at the top of the array, so removal is quick and easy; the item is removed and the Front arrow moves down to point to the new top of the array. No shifting or comparisons are necessary.

In the PriorityQ Workshop applet, we show Front and Rear arrows to provide a comparison with an ordinary queue, but they’re not really necessary. The algorithms

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know that the front of the queue is always at the top of the array at nItems-1, and they insert items in order, not at the rear. Figure 4.12 shows the operation of the PriorityQ class methods.

500

43

Front

43

Front

109

109

320

320

500

632

632

841

841

Rear

Rear

New item inserted in priority queue

43

43

Front

109

109

109

Front

320

320

320

Front

500

500

500

632

632

632

841

Rear

841

Rear

841

Rear

Two items removed from front of priority queue

**FIGURE 4.12 **

**The Peek and New Buttons**

You can peek at the minimum item (find its value without removing it) with the Peek button, and you can create a new, empty, priority queue with the New button.

**Other Implementation Possibilities**

The implementation shown in the PriorityQ Workshop applet isn’t very efficient for insertion, which involves moving an average of half the items.

Another approach, which also uses an array, makes no attempt to keep the items in sorted order. New items are simply inserted at the top of the array. This makes

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insertion very quick, but unfortunately it makes deletion slow because the smallest item must be searched for. This approach requires examining all the items and shifting half of them, on the average, down to fill in the hole. In most situations the quick-deletion approach shown in the Workshop applet is preferred.

For small numbers of items, or situations in which speed isn’t critical, implementing a priority queue with an array is satisfactory. For larger numbers of items, or when speed is critical, the heap is a better choice.

**Java Code for a Priority Queue**

The Java code for a simple array-based priority queue is shown in Listing 4.6.

**LISTING 4.6**

The priorityQ.java Program

// priorityQ.java

// demonstrates priority queue

// to run this program: C>java PriorityQApp

////////////////////////////////////////////////////////////////

class PriorityQ

{

// array in sorted order, from max at 0 to min at size-1

private int maxSize;

private long[] queArray;

private int nItems;

//-------------------------------------------------------------

public PriorityQ(int s) // constructor

{

maxSize = s;

queArray = new long[maxSize];

nItems = 0;

}

//-------------------------------------------------------------

public void insert(long item) // insert item

{

int j;

if(nItems==0) // if no items, queArray[nItems++] = item; // insert at 0

else // if items,

{

for(j=nItems-1; j>=0; j--) // start at end,

{

if( item > queArray[j] ) // if new item larger, queArray[j+1] = queArray[j]; // shift upward

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**LISTING 4.6**

Continued

else // if smaller,

break; // done shifting

} // end for

queArray[j+1] = item; // insert it

nItems++;

} // end else (nItems > 0)

} // end insert()

//-------------------------------------------------------------

public long remove() // remove minimum item

{ return queArray[--nItems]; }

//-------------------------------------------------------------

public long peekMin() // peek at minimum item

{ return queArray[nItems-1]; }

//-------------------------------------------------------------

public boolean isEmpty() // true if queue is empty

{ return (nItems==0); }

//-------------------------------------------------------------

public boolean isFull() // true if queue is full

{ return (nItems == maxSize); }

//-------------------------------------------------------------

} // end class PriorityQ

////////////////////////////////////////////////////////////////

class PriorityQApp

{

public static void main(String[] args) throws IOException

{

PriorityQ thePQ = new PriorityQ(5);

thePQ.insert(30);

thePQ.insert(50);

thePQ.insert(10);

thePQ.insert(40);

thePQ.insert(20);

while( !thePQ.isEmpty() )

{

long item = thePQ.remove();

System.out.print(item + “ “); // 10, 20, 30, 40, 50

} // end while

System.out.println(“”);

} // end main()

//-------------------------------------------------------------

} // end class PriorityQApp

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In main() we insert five items in random order, and then remove and display them.

The smallest item is always removed first, so the output is 10, 20, 30, 40, 50

The insert() method checks whether there are any items; if not, it inserts one at index 0. Otherwise, it starts at the top of the array and shifts existing items upward until it finds the place where the new item should go. Then it inserts the item and increments nItems. Note that if there’s any chance the priority queue is full, you should check for this possibility with isFull() before using insert().

The front and rear fields aren’t necessary as they were in the Queue class because, as we noted, front is always at nItems-1 and rear is always at 0.

The remove() method is simplicity itself: It decrements nItems and returns the item from the top of the array. The peekMin() method is similar, except it doesn’t decrement nItems. The isEmpty() and isFull() methods check if nItems is 0 or maxSize, respectively.

**Efficiency of Priority Queues**

In the priority-queue implementation we show here, insertion runs in O(N) time, while deletion takes O(1) time. We’ll see how to improve insertion time with heaps in Chapter 12.

**Parsing Arithmetic Expressions**

So far in this chapter, we’ve introduced three different data storage structures. Let’s shift gears now and focus on an important application for one of these structures.

This application is

In some sense this section should be considered optional. It’s not a prerequisite to the rest of the book, and writing code to parse arithmetic expressions is probably not something you need to do every day, unless you are a compiler writer or are designing pocket calculators. Also, the coding details are more complex than any we’ve seen so far. However, seeing this important use of stacks is educational, and the issues raised are interesting in their own right.

As it turns out, it’s fairly difficult, at least for a computer algorithm, to evaluate an arithmetic expression directly. It’s easier for the algorithm to use a two-step process:

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**1. **Transform the arithmetic expression into a different format, called postfix notation.

**2. **Evaluate the postfix expression.

Step 1 is a bit involved, but step 2 is easy. In any case, this two-step approach results in a simpler algorithm than trying to parse the arithmetic expression directly. Of course, for a human it’s easier to parse the ordinary arithmetic expression. We’ll return to the difference between the human and computer approaches in a moment.

Before we delve into the details of steps 1 and 2, we’ll introduce postfix notation.

**Postfix Notation**

Everyday arithmetic expressions are written with an

and 4⁄7, or, using letters to stand for numbers, A+B and A⁄B.

In postfix notation (which is also called Reverse Polish Notation, or RPN, because it was invented by a Polish mathematician), the operator

Thus, A+B becomes AB+, and A⁄B becomes AB/. More complex infix expressions can likewise be translated into postfix notation, as shown in Table 4.2. We’ll explain how the postfix expressions are generated in a moment.

**TABLE 4.2**

Infix and Postfix Expressions

**Infix**

**Postfix**

A+B–C

AB+C–

A*B/C

AB*C/

A+B*C

ABC*+

A*B+C

AB*C+

A*(B+C)

ABC+*

A*B+C*D

AB*CD*+

(A+B)*(C–D)

AB+CD–*

((A+B)*C)–D

AB+C*D–

A+B*(C–D/(E+F))

ABCDEF+/–*+

Some computer languages also have an operator for raising a quantity to a power (typically, the ^ character), but we’ll ignore that possibility in this discussion.

Besides infix and postfix, there’s also a

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**Translating Infix to Postfix**

The next several pages are devoted to explaining how to translate an expression from infix notation into postfix. This algorithm is fairly involved, so don’t worry if every detail isn’t clear at first. If you get bogged down, you may want to skip ahead to the section “Evaluating Postfix Expressions.” To understand how to create a postfix expression, you might find it helpful to see how a postfix expression is evaluated; for example, how the value 14 is extracted from the expression 234+*, which is the postfix equivalent of 2*(3+4). (Notice that in this discussion, for ease of writing, we restrict ourselves to expressions with single-digit numbers, although these expressions may evaluate to multidigit numbers.)

**How Humans Evaluate Infix**

How do you translate infix to postfix? Let’s examine a slightly easier question first: How does a human evaluate a normal infix expression? Although, as we stated earlier, such evaluation is difficult for a computer, we humans do it fairly easily because of countless hours in Mr. Klemmer’s math class. It’s not hard for us to find the answer to 3+4+5, or 3*(4+5). By analyzing how we evaluate this expression, we can achieve some insight into the translation of such expressions into postfix.

Roughly speaking, when you “solve” an arithmetic expression, you follow rules something like this:

**1. **You read from left to right. (At least, we’ll assume this is true. Sometimes people skip ahead, but for purposes of this discussion, you should assume you must read methodically, starting at the left.)

**2. **When you’ve read enough to evaluate two operands and an operator, you do the calculation and substitute the answer for these two operands and operator.

(You may also need to solve other pending operations on the left, as we’ll see later.)

**3. **You continue this process—going from left to right and evaluating when possible—until the end of the expression.

Tables 4.3, 4.4, and 4.5 show three examples of how simple infix expressions are evaluated. Later, in Tables 4.6, 4.7, and 4.8, we’ll see how closely these evaluations mirror the process of translating infix to postfix.

To evaluate 3+4–5, you would carry out the steps shown in Table 4.3.

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**TABLE 4.3**

Evaluating 3+4–5

**Item Read**

**Expression Parsed So Far**

**Comments**

3

3

+

3+

4

3+4

–

7

When you see the –, you can evaluate 3+4.

7–

5

7–5

End

2

When you reach the end of the expression, you

can evaluate 7–5.

You can’t evaluate the 3+4 until you see what operator follows the 4. If it’s an * or /, you need to wait before applying the + sign until you’ve evaluated the * or /.

However, in this example the operator following the 4 is a –, which has the same precedence as a +, so when you see the –, you know you can evaluate 3+4, which is 7. The 7 then replaces the 3+4. You can evaluate the 7–5 when you arrive at the end of the expression.

Figure 4.13 shows this process in more detail. Notice how you go from left to right reading items from the input, and then, when you have enough information, you go from right to left, recalling previously examined input and evaluating each operand-operator-operand combination.

Because of precedence relationships, evaluating 3+4*5 is a bit more complicated, as shown in Table 4.4.

**TABLE 4.4**

Evaluating 3+4*5

**Item Read**

**Expression Parsed So Far**

**Comments**

3

3

+

3+

4

3+4

*

3+4*

You can’t evaluate 3+4 because * is higher

precedence than +.

5

3+4*5

When you see the 5, you can evaluate 4*5.

3+20

End

23

When you see the end of the expression, you can

evaluate 3+20.

Here you can’t add the 3 until you know the result of 4*5. Why not? Because multiplication has a higher precedence than addition. In fact, both * and / have a higher precedence than + and –, so all multiplications and divisions must be carried out before any additions or subtractions (unless parentheses dictate otherwise; see the next example).

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1 Read

2 Read

3 Read

4 Read

the 3

the +

the 4

the -

3 + 4 - 5 End

8 Evaluate

7 Recall

6 Recall

5 Recall

3+4

the 3

the +

the 4

9 Read

10 Read

the 5

the End

7 - 5 End

14 Evaluate 13 Recall

12 Recall

11 Recall

7-5

the 7

the -

the 5

2

15 Recall

the 2

**FIGURE 4.13 **

Often you can evaluate as you go from left to right, as in the preceding example.

However, you need to be sure, when you come to an operand-operator-operand combination such as A+B, that the operator on the right side of the B isn’t one with a higher precedence than the +. If it does have a higher precedence, as in this example, you can’t do the addition yet. However, after you’ve read the 5, the multiplication can be carried out because it has the highest priority; it doesn’t matter whether a * or / follows the 5. However, you still can’t do the addition until you’ve found out what’s beyond the 5. When you find there’s nothing beyond the 5 but the end of the expression, you can go ahead and do the addition. Figure 4.14 shows this process.

Parentheses are used to override the normal precedence of operators. Table 4.5 shows how you would evaluate 3*(4+5). Without the parentheses, you would do the multiplication first; with them, you do the addition first.

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1 Read

2 Read

3 Read

4 Read

5 Read

the 3

the +

the 4

the *

the 5

3 + 4 * 5 End

9 Evaluate

8 Recall

7 Recall

6 Recall

4*5

the 4

the *

the 5

10 Read

the End

3 + 20 End

14 Evaluate 13 Recall

12 Recall

11 Recall

3+20

the 3

the +

the 20

23

15 Recall

the 23

**FIGURE 4.14 **

**TABLE 4.5**

Evaluating 3*(4+5)

**Item Read**

**Expression Parsed So Far**

**Comments**

3

3

*

3*

(

3*(

4

3*(4

You can’t evaluate 3*4 because of the parenthesis.

+

3*(4+

5

3*(4+5

You can’t evaluate 4+5 yet.

)

3*(4+5)

When you see the ), you can evaluate 4+5.

3*9

After you’ve evaluated 4+5, you can evaluate 3*9.

27

End

Nothing left to evaluate.

Here we can’t evaluate anything until we’ve reached the closing parenthesis.

Multiplication has a higher or equal precedence compared to the other operators, so ordinarily we could carry out 3*4 as soon as we see the 4. However, parentheses have an even higher precedence than * and /. Accordingly, we must evaluate anything in parentheses before using the result as an operand in any other calculation. The

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closing parenthesis tells us we can go ahead and do the addition. We find that 4+5 is 9, and when we know this, we can evaluate 3*9 to obtain 27. Reaching the end of the expression is an anticlimax because there’s nothing left to evaluate. This process is shown in Figure 4.15.

1 Read

2 Read

3 Read

4 Read

5 Read

6 Read

7 Read

the 3

the *

the (

the 4

the +

the 5

the )

3 * ( 4 + 5 ) End

13 Evaluate 12 Discard

11 Recall

10 Recall

9 Recall

8 Discard

4+5

the (

the 4

the +

the 5

the )

3 * 9 End

17 Evaluate 16 Recall

15 Recall

14 Recall

3*9

the 3

the *

the 9

18 Recall

the End

27 End

19 Recall

the 27

**FIGURE 4.15 **

As we’ve seen, in evaluating an infix arithmetic expression, you go both forward and backward through the expression. You go forward (left to right) reading operands and operators. When you have enough information to apply an operator, you go backward, recalling two operands and an operator and carrying out the arithmetic.

Sometimes you must defer applying operators if they’re followed by higher precedence operators or by parentheses. When this happens, you must apply the later, higher-precedence, operator first; then go backward (to the left) and apply earlier operators.

We could write an algorithm to carry out this kind of evaluation directly. However, as we noted, it’s actually easier to translate into postfix notation first.

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**How Humans Translate Infix to Postfix**

To translate infix to postfix notation, you follow a similar set of rules to those for evaluating infix. However, there are a few small changes. You don’t do any arithmetic. The idea is not to evaluate the infix expression, but to rearrange the operators and operands into a different format: postfix notation. The resulting postfix expression will be evaluated later.

As before, you read the infix from left to right, looking at each character in turn. As you go along, you copy these operands and operators to the postfix output string.

The trick is knowing when to copy what.

If the character in the infix string is an operand, you copy it immediately to the postfix string. That is, if you see an A in the infix, you write an A to the postfix.

There’s never any delay: You copy the operands as you get to them, no matter how long you must wait to copy their associated operators.

Knowing when to copy an operator is more complicated, but it’s the same as the rule for evaluating infix expressions. Whenever you could have used the operator to evaluate part of the infix expression (if you were evaluating instead of translating to postfix), you instead copy it to the postfix string.

Table 4.6 shows how A+B–C is translated into postfix notation.

**TABLE 4.6**

Translating A+B–C into Postfix

**Character**

**Infix**

**Postfix**

**Comments**

**Read from**

**Expression**

**Expression**

**Infix**

**Parsed So**

**Written So**

**Expression**

**Far**

**Far**

A

A

A

+

A+

A

B

A+B

AB

–

A+B–

AB+

When you see the –, you can copy the +

to the postfix string.

C

A+B–C

AB+C

End

A+B–C

AB+C–

When you reach the end of the expression,

you can copy the –.

Notice the similarity of this table to Table 4.3, which showed the evaluation of the infix expression 3+4–5. At each point where you would have done an evaluation in the earlier table, you instead simply write an operator to the postfix output.

Table 4.7 shows the translation of A+B*C to postfix. This evaluation is similar to Table 4.4, which covered the evaluation of 3+4*5.

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**TABLE 4.7**

Translating A+B*C to Postfix

**Character**

**Infix**

**Postfix**

**Comments**

**Read from**

**Expression**

**Expression**

**Infix**

**Parsed So**

**Written So**

**Expression**

**Far**

**Far**

A

A

A

+

A+

A

B

A+B

AB

*

A+B*

AB

You can’t copy the + because * is

higher precedence than +.

C

A+B*C

ABC

When you see the C, you can copy the *.

A+B*C

ABC*

End

A+B*C

ABC*+

When you see the end of the

expression, you can copy the +.

As the final example, Table 4.8 shows how A*(B+C) is translated to postfix. This process is similar to evaluating 3*(4+5) in Table 4.5. You can’t write any postfix operators until you see the closing parenthesis in the input.

**TABLE 4.8**

Translating A*(B+C) into Postfix

**Character**

**Infix**

**Postfix**

**Comments**

**Read from**

**Expression**

**Expression**

**Infix**

**Parsed so**

**Written So**

**Expression**

**Far**

**Far**

A

A

A

*

A*

A

(

A*(

A

B

A*(B

AB

You can’t copy * because of the

parenthesis.

+

A*(B+

AB

C

A*(B+C

ABC

You can’t copy the + yet.

)

A*(B+C)

ABC+

When you see the ), you can copy the +.

A*(B+C)

ABC+*

After you’ve copied the +, you can copy

the *.

End

A*(B+C)

ABC+*

Nothing left to copy.

As in the numerical evaluation process, you go both forward and backward through the infix expression to complete the translation to postfix. You can’t write an operator to the output (postfix) string if it’s followed by a higher-precedence operator or a left parenthesis. If it is, the higher-precedence operator or the operator in parentheses must be written to the postfix before the lower-priority operator.

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**Saving Operators on a Stack**

You’ll notice in both Table 4.7 and Table 4.8 that the order of the operators is reversed going from infix to postfix. Because the first operator can’t be copied to the output until the second one has been copied, the operators were output to the postfix string in the opposite order they were read from the infix string. A longer example may make this operation clearer. Table 4.9 shows the translation to postfix of the infix expression A+B*(C–D). We include a column for stack contents, which we’ll explain in a moment.

**TABLE 4.9**

Translating A+B*(C–D) to Postfix

**Character**

**Infix**

**Postfix**

**Stack**

**Read from**

**Expression**

**Expression**

**Contents**

**Infix**

**Parsed So**

**Written So**

**Expression**

**Far**

**Far**

A

A

A

+

A+

A

+

B

A+B

AB

+

*

A+B*

AB

+*

(

A+B*(

AB

+*(

C

A+B*(C

ABC

+*(

–

A+B*(C–

ABC

+*(–

D

A+B*(C–D

ABCD

+*(–

)

A+B*(C–D)

ABCD–

+*(

A+B*(C–D)

ABCD–

+*(

A+B*(C–D)

ABCD–

+*

A+B*(C–D)

ABCD–*

+

A+B*(C–D)

ABCD–*+

Here we see the order of the operands is +*– in the original infix expression, but the reverse order, –*+, in the final postfix expression. This happens because * has higher precedence than +, and –, because it’s in parentheses, has higher precedence than *.

This order reversal suggests a stack might be a good place to store the operators while we’re waiting to use them. The last column in Table 4.9 shows the stack contents at various stages in the translation process.

Popping items from the stack allows you to, in a sense, go backward (right to left) through the input string. You’re not really examining the entire input string, only the operators and parentheses. They were pushed on the stack when reading the input, so now you can recall them in reverse order by popping them off the stack.

The operands (A, B, and so on) appear in the same order in infix and postfix, so you can write each one to the output as soon as you encounter it; they don’t need to be stored on a stack.

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**Translation Rules**

Let’s make the rules for infix-to-postfix translation more explicit. You read items from the infix input string and take the actions shown in Table 4.10. These actions are described in pseudocode, a blend of Java and English.

In this table, the < and >= symbols refer to the operator precedence relationship, not numerical values. The opThis operator has just been read from the infix input, while the opTop operator has just been popped off the stack.

**TABLE 4.10**

Infix to Postfix Translation Rules

**Item Read from Input**

**Action**

**(Infix)**

Operand

Write it to output (postfix)

Open parenthesis (

Push it on stack

Close parenthesis )

While stack not empty, repeat the following:

Pop an item,

If item is not (, write it to output

Quit loop if item is (

Operator (opThis)

If stack empty,

Push opThis

Otherwise,

While stack not empty, repeat:

Pop an item,

If item is (, push it, or

If item is an operator (opTop), and

If opTop < opThis, push opTop, or

If opTop >= opThis, output opTop

Quit loop if opTop < opThis or item is (

Push opThis

No more items

While stack not empty,

Pop item, output it.

Convincing yourself that these rules work may take some effort. Tables 4.11, 4.12, and 4.13 show how the rules apply to three example infix expressions. These tables are similar to Tables 4.6, 4.7, and 4.8, except that the relevant rules for each step have been added. Try creating similar tables by starting with other simple infix expressions and using the rules to translate some of them to postfix.

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**TABLE 4.11**

Translation Rules Applied to A+B–C

**Character**

**Infix**

**Postfix**

**Stack**

**Rule**

**Read from**

**Parsed So**

**Written So**

**Contents**

**Infix**

**Far**

**Far**

A

A

A

Write operand to output.

+

A+

A

+

If stack empty, push opThis.

B

A+B

AB

+

Write operand to output.

–

A+B–

AB

Stack not empty, so pop item.

A+B–

AB+

opThis is –, opTop is +,

opTop>=opThis, so output

opTop.

A+B–

AB+

–

Then push opThis.

C

A+B–C

AB+C

–

Write operand to output.

End

A+B–C

AB+C–

Pop leftover item, output it.

**TABLE 4.12**

Translation Rules Applied to A+B*C

**Character**

**Infix**

**Postfix**

**Stack**

**Rule**

**Read From**

**Parsed**

**Written**

**Contents**

**Infix So **

**Far**

**So **

**Far**

A

A

A

Write operand to postfix.

+

A+

A

+

If stack empty, push opThis.

B

A+B

AB

+

Write operand to output.

*

A+B*

AB

+

Stack not empty, so pop

opTop.

A+B*

AB

+

opThis is *, opTop is +,

opTop<opThis, so push

opTop.

A+B*

AB

+*

Then push opThis.

C

A+B*C

ABC

+*

Write operand to output.

End

A+B*C

ABC*

+

Pop leftover item, output it.

A+B*C

ABC*+

Pop leftover item, output it.

**TABLE 4.13**

Translation Rules Applied to A*(B+C)

**Character**

**Infix**

**Postfix**

**Stack**

**Rule**

**Read From**

**Parsed**

**Written**

**Contents**

**Infix**

**So Far**

**So Far**

A

A

A

Write operand to postfix.

*

A*

A

*

If stack empty, push opThis.

(

A*(

A

*(

Push ( on stack.

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161

**TABLE 4.13**

Continued

**Character**

**Infix**

**Postfix**

**Stack**

**Rule**

**Read From**

**Parsed**

**Written**

**Contents**

**Infix**

**So Far**

**So Far**

B

A*(B

AB

*(

Write operand to postfix.

+

A*(B+

AB

*

Stack not empty, so pop item.

A*(B+

AB

*(

It’s (, so push it.

A*(B+

AB

*(+

Then push opThis.

C

A*(B+C

ABC

*(+

Write operand to postfix.

)

A*(B+C)

ABC+

*(

Pop item, write to output.

A*(B+C)

ABC+

*

Quit popping if (.

End

A*(B+C)

ABC+*

Pop leftover item, output it.

**Java Code to Convert Infix to Postfix**

Listing 4.7 shows the infix.java program, which uses the rules from Table 4.10 to translate an infix expression to a postfix expression.

**LISTING 4.7**

The infix.java Program

// infix.java

// converts infix arithmetic expressions to postfix

// to run this program: C>java InfixApp

import java.io.*; // for I/O

////////////////////////////////////////////////////////////////

class StackX

{

private int maxSize;

private char[] stackArray;

private int top;

//--------------------------------------------------------------

public StackX(int s) // constructor

{

maxSize = s;

stackArray = new char[maxSize];

top = -1;

}

//--------------------------------------------------------------

public void push(char j) // put item on top of stack

{ stackArray[++top] = j; }

//--------------------------------------------------------------

public char pop() // take item from top of stack

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**LISTING 4.7**

Continued

{ return stackArray[top--]; }

//--------------------------------------------------------------

public char peek() // peek at top of stack

{ return stackArray[top]; }

//--------------------------------------------------------------

public boolean isEmpty() // true if stack is empty

{ return (top == -1); }

//-------------------------------------------------------------

public int size() // return size

{ return top+1; }

//--------------------------------------------------------------

public char peekN(int n) // return item at index n

{ return stackArray[n]; }

//--------------------------------------------------------------

public void displayStack(String s)

{

System.out.print(s);

System.out.print(“Stack (bottom-->top): “);

for(int j=0; j<size(); j++)

{

System.out.print( peekN(j) );

System.out.print(‘ ‘);

}

System.out.println(“”);

}

//--------------------------------------------------------------

} // end class StackX

////////////////////////////////////////////////////////////////

class InToPost // infix to postfix conversion

{

private StackX theStack;

private String input;

private String output = “”;

//--------------------------------------------------------------

public InToPost(String in) // constructor

{

input = in;

int stackSize = input.length();

theStack = new StackX(stackSize);

}

//--------------------------------------------------------------

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**LISTING 4.7**

Continued

public String doTrans() // do translation to postfix

{

for(int j=0; j<input.length(); j++)

{

char ch = input.charAt(j);

theStack.displayStack(“For “+ch+” “); // *diagnostic*

switch(ch)

{

case ‘+’: // it’s + or -

case ‘-’:

gotOper(ch, 1); // go pop operators

break; // (precedence 1)

case ‘*’: // it’s * or /

case ‘/’:

gotOper(ch, 2); // go pop operators

break; // (precedence 2)

case ‘(‘: // it’s a left paren

theStack.push(ch); // push it

break;

case ‘)’: // it’s a right paren

gotParen(ch); // go pop operators

break;

default: // must be an operand

output = output + ch; // write it to output

break;

} // end switch

} // end for

while( !theStack.isEmpty() ) // pop remaining opers

{

theStack.displayStack(“While “); // *diagnostic*

output = output + theStack.pop(); // write to output

}

theStack.displayStack(“End “); // *diagnostic*

return output; // return postfix

} // end doTrans()

//--------------------------------------------------------------

public void gotOper(char opThis, int prec1)

{ // got operator from input while( !theStack.isEmpty() )

{

char opTop = theStack.pop();

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**LISTING 4.7**

Continued

if( opTop == ‘(‘ ) // if it’s a ‘(‘

{

theStack.push(opTop); // restore ‘(‘

break;

}

else // it’s an operator

{

int prec2; // precedence of new op

if(opTop==’+’ || opTop==’-’) // find new op prec

prec2 = 1;

else

prec2 = 2;

if(prec2 < prec1) // if prec of new op less

{ // than prec of old

theStack.push(opTop); // save newly-popped op

break;

}

else // prec of new not less

output = output + opTop; // than prec of old

} // end else (it’s an operator)

} // end while

theStack.push(opThis); // push new operator

} // end gotOp()

//--------------------------------------------------------------

public void gotParen(char ch)

{ // got right paren from input while( !theStack.isEmpty() )

{

char chx = theStack.pop();

if( chx == ‘(‘ ) // if popped ‘(‘

break; // we’re done

else // if popped operator

output = output + chx; // output it

} // end while

} // end popOps()

//--------------------------------------------------------------

} // end class InToPost

////////////////////////////////////////////////////////////////

class InfixApp

{

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**LISTING 4.7**

Continued

public static void main(String[] args) throws IOException

{

String input, output;

while(true)

{

System.out.print(“Enter infix: “);

System.out.flush();

input = getString(); // read a string from kbd if( input.equals(“”) ) // quit if [Enter]

break;

// make a translator

InToPost theTrans = new InToPost(input);

output = theTrans.doTrans(); // do the translation

System.out.println(“Postfix is “ + output + ‘\n’);

} // end while

} // end main()

//--------------------------------------------------------------

public static String getString() throws IOException

{

String s = br.readLine();

return s;

}

//--------------------------------------------------------------

} // end class InfixApp

////////////////////////////////////////////////////////////////

The main() routine in the InfixApp class asks the user to enter an infix expression.

The input is read with the readString() utility method. The program creates an InToPost object, initialized with the input string. Then it calls the doTrans() method for this object to perform the translation. This method returns the postfix output string, which is displayed.

The doTrans() method uses a switch statement to handle the various translation rules shown in Table 4.10. It calls the gotOper() method when it reads an operator and the gotParen() method when it reads a closing parenthesis, ). These methods implement the second two rules in the table, which are more complex than other rules.

We’ve included a displayStack() method to display the entire contents of the stack in the StackX class. In theory, this isn’t playing by the rules; you’re supposed to access the item only at the top. However, as a diagnostic aid, this routine is useful if

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you want to see the contents of the stack at each stage of the translation. Here’s some sample interaction with infix.java:

Enter infix: A*(B+C)-D/(E+F)

For A Stack (bottom-->top):

For * Stack (bottom-->top):

For ( Stack (bottom-->top): *

For B Stack (bottom-->top): * (

For + Stack (bottom-->top): * (

For C Stack (bottom-->top): * ( +

For ) Stack (bottom-->top): * ( +

For - Stack (bottom-->top): *

For D Stack (bottom-->top): -

For / Stack (bottom-->top): -

For ( Stack (bottom-->top): - /

For E Stack (bottom-->top): - / (

For + Stack (bottom-->top): - / (

For F Stack (bottom-->top): - / ( +

For ) Stack (bottom-->top): - / ( +

While Stack (bottom-->top): - /

While Stack (bottom-->top): -

End Stack (bottom-->top):

Postfix is ABC+*DEF+/-

The output shows where the displayStack() method was called (from the for loop, the while loop, or at the end of the program) and, within the for loop, what character has just been read from the input string.

You can use single-digit numbers like 3 and 7 instead of symbols like A and B.

They’re all just characters to the program. For example: Enter infix: 2+3*4

For 2 Stack (bottom-->top):

For + Stack (bottom-->top):

For 3 Stack (bottom-->top): +

For * Stack (bottom-->top): +

For 4 Stack (bottom-->top): + *

While Stack (bottom-->top): + *

While Stack (bottom-->top): +

End Stack (bottom-->top):

Postfix is 234*+

Of course, in the postfix output, the 234 means the separate numbers 2, 3, and 4.

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The infix.java program doesn’t check the input for errors. If you type an incorrect infix expression, the program will provide erroneous output or crash and burn.

Experiment with this program. Start with some simple infix expressions, and see if you can predict what the postfix will be. Then run the program to verify your answer. Pretty soon, you’ll be a postfix guru, much sought after at cocktail parties.

**Evaluating Postfix Expressions**

As you can see, converting infix expressions to postfix expressions is not trivial. Is all this trouble really necessary? Yes, the payoff comes when you evaluate a postfix expression. Before we show how simple the algorithm is, let’s examine how a human might carry out such an evaluation.

**How Humans Evaluate Postfix**

Figure 4.16 shows how a human can evaluate a postfix expression using visual inspection and a pencil.

❺

❹

❷

❶

❸

3 4 5 + * 6 1 2 + / -

9

3

27

2

25

**FIGURE 4.16 **

Start with the first operator on the left, and draw a circle around it and the two operands to its immediate left. Then apply the operator to these two operands—

performing the actual arithmetic—and write down the result inside the circle. In the figure, evaluating 4+5 gives 9.

Now go to the next operator to the right, and draw a circle around it, the circle you already drew, and the operand to the left of that. Apply the operator to the previous circle and the new operand, and write the result in the new circle. Here 3*9 gives 27.

Continue this process until all the operators have been applied: 1+2 is 3, and 6/3 is 2. The answer is the result in the largest circle: 27–2 is 25.

**Rules for Postfix Evaluation**

How do we write a program to reproduce this evaluation process? As you can see, each time you come to an operator, you apply it to the last two operands you’ve seen. This suggests that it might be appropriate to store the operands on a stack.

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(This is the opposite of the infix-to-postfix translation algorithm, where

**TABLE 4.14**

Evaluating a Postfix Expression

**Item Read from Postfix**

**Action**

**Expression**

Operand

Push it onto the stack.

Operator

Pop the top two operands from the stack and apply the operator to them. Push the result.

When you’re done, pop the stack to obtain the answer. That’s all there is to it. This process is the computer equivalent of the human circle-drawing approach of Figure 4.16.

**Java Code to Evaluate Postfix Expressions**

In the infix-to-postfix translation, we used symbols (A, B, and so on) to stand for numbers. This approach worked because we weren’t performing arithmetic operations on the operands but merely rewriting them in a different format.

Now we want to evaluate a postfix expression, which means carrying out the arithmetic and obtaining an answer. Thus, the input must consist of actual numbers. To simplify the coding, we’ve restricted the input to single-digit numbers.

Our program evaluates a postfix expression and outputs the result. Remember numbers are restricted to one digit. Here’s some simple interaction: Enter postfix: 57+

5 Stack (bottom-->top):

7 Stack (bottom-->top): 5

+ Stack (bottom-->top): 5 7

Evaluates to 12

You enter digits and operators, with no spaces. The program finds the numerical equivalent. Although the input is restricted to single-digit numbers, the results are not; it doesn’t matter if something evaluates to numbers greater than 9. As in the infix.java program, we use the displayStack() method to show the stack contents at each step. Listing 4.8 shows the postfix.java program.

**LISTING 4.8**

The postfix.java Program

// postfix.java

// parses postfix arithmetic expressions

// to run this program: C>java PostfixApp

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**LISTING 4.8**

Continued

import java.io.*; // for I/O

////////////////////////////////////////////////////////////////

class StackX

{

private int maxSize;

private int[] stackArray;

private int top;

//--------------------------------------------------------------

public StackX(int size) // constructor

{

maxSize = size;

stackArray = new int[maxSize];

top = -1;

}

//--------------------------------------------------------------

public void push(int j) // put item on top of stack

{ stackArray[++top] = j; }

//--------------------------------------------------------------

public int pop() // take item from top of stack

{ return stackArray[top--]; }

//--------------------------------------------------------------

public int peek() // peek at top of stack

{ return stackArray[top]; }

//--------------------------------------------------------------

public boolean isEmpty() // true if stack is empty

{ return (top == -1); }

//--------------------------------------------------------------

public boolean isFull() // true if stack is full

{ return (top == maxSize-1); }

//--------------------------------------------------------------

public int size() // return size

{ return top+1; }

//--------------------------------------------------------------

public int peekN(int n) // peek at index n

{ return stackArray[n]; }

//--------------------------------------------------------------

public void displayStack(String s)

{

System.out.print(s);

System.out.print(“Stack (bottom-->top): “);

for(int j=0; j<size(); j++)

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**LISTING 4.8**

Continued

{

System.out.print( peekN(j) );

System.out.print(‘ ‘);

}

System.out.println(“”);

}

//--------------------------------------------------------------

} // end class StackX

////////////////////////////////////////////////////////////////

class ParsePost

{

private StackX theStack;

private String input;

//--------------------------------------------------------------

public ParsePost(String s)

{ input = s; }

//--------------------------------------------------------------

public int doParse()

{

theStack = new StackX(20); // make new stack char ch;

int j;

int num1, num2, interAns;

for(j=0; j<input.length(); j++) // for each char,

{

ch = input.charAt(j); // read from input theStack.displayStack(“”+ch+” “); // *diagnostic*

if(ch >= ‘0’ && ch <= ‘9’) // if it’s a number theStack.push( (int)(ch-’0’) ); // push it

else // it’s an operator

{

num2 = theStack.pop(); // pop operands

num1 = theStack.pop();

switch(ch) // do arithmetic

{

case ‘+’:

interAns = num1 + num2;

break;

case ‘-’:

interAns = num1 - num2;

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**LISTING 4.8**

Continued

break;

case ‘*’:

interAns = num1 * num2;

break;

case ‘/’:

interAns = num1 / num2;

break;

default:

interAns = 0;

} // end switch

theStack.push(interAns); // push result

} // end else

} // end for

interAns = theStack.pop(); // get answer return interAns;

} // end doParse()

} // end class ParsePost

////////////////////////////////////////////////////////////////

class PostfixApp

{

public static void main(String[] args) throws IOException

{

String input;

int output;

while(true)

{

System.out.print(“Enter postfix: “);

System.out.flush();

input = getString(); // read a string from kbd if( input.equals(“”) ) // quit if [Enter]

break;

// make a parser

ParsePost aParser = new ParsePost(input);

output = aParser.doParse(); // do the evaluation

System.out.println(“Evaluates to “ + output);

} // end while

} // end main()

//--------------------------------------------------------------

public static String getString() throws IOException

{

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**LISTING 4.8**

Continued

String s = br.readLine();

return s;

}

//--------------------------------------------------------------

} // end class PostfixApp

////////////////////////////////////////////////////////////////

The main() method in the PostfixApp class gets the postfix string from the user and then creates a ParsePost object, initialized with this string. It then calls the doParse() method of ParsePost to carry out the evaluation.

The doParse() method reads through the input string character by character. If the character is a digit, it’s pushed onto the stack. If it’s an operator, it’s applied immediately to the two operators on the top of the stack. (These operators are guaranteed to be on the stack already because the input string is in postfix notation.) The result of the arithmetic operation is pushed onto the stack. After the last character (which must be an operator) is read and applied, the stack contains only one item, which is the answer to the entire expression.

Here’s some interaction with more complex input: the postfix expression 345+*612+/–, which we showed a human evaluating in Figure 4.16. This expression corresponds to the infix 3*(4+5)–6/(1+2). (We saw an equivalent translation using letters instead of numbers in the previous section: A*(B+C)–D/(E+F) in infix is ABC+*DEF+/– in postfix.) Here’s how the postfix is evaluated by the postfix.java program:

Enter postfix: 345+*612+/-

3 Stack (bottom-->top):

4 Stack (bottom-->top): 3

5 Stack (bottom-->top): 3 4

+ Stack (bottom-->top): 3 4 5

* Stack (bottom-->top): 3 9

6 Stack (bottom-->top): 27

1 Stack (bottom-->top): 27 6

2 Stack (bottom-->top): 27 6 1

+ Stack (bottom-->top): 27 6 1 2

/ Stack (bottom-->top): 27 6 3

- Stack (bottom-->top): 27 2

Evaluates to 25

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173

As with the infix.java program (Listing 4.7), postfix.java doesn’t check for input errors. If you type in a postfix expression that doesn’t make sense, results are unpredictable.

Experiment with the program. Trying different postfix expressions and seeing how they’re evaluated will give you an understanding of the process faster than reading about it.

**Summary**

• Stacks, queues, and priority queues are data structures usually used to simplify certain programming operations.

• In these data structures, only one data item can be accessed.

• A stack allows access to the last item inserted.

• The important stack operations are pushing (inserting) an item onto the top of the stack and popping (removing) the item that’s on the top.

• A queue allows access to the first item that was inserted.

• The important queue operations are inserting an item at the rear of the queue and removing the item from the front of the queue.

• A queue can be implemented as a circular queue, which is based on an array in which the indices wrap around from the end of the array to the beginning.

• A priority queue allows access to the smallest (or sometimes the largest) item.

• The important priority queue operations are inserting an item in sorted order and removing the item with the smallest key.

• These data structures can be implemented with arrays or with other mechanisms such as linked lists.

• Ordinary arithmetic expressions are written in infix notation, so-called because the operator is written between the two operands.

• In postfix notation, the operator follows the two operands.

• Arithmetic expressions are typically evaluated by translating them to postfix notation and then evaluating the postfix expression.

• A stack is a useful tool both for translating an infix to a postfix expression and for evaluating a postfix expression.

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**Questions**

These questions are intended as a self-test for readers. Answers may be found in Appendix C.

**1. **Suppose you push 10, 20, 30, and 40 onto the stack. Then you pop three items.

Which one is left on the stack?

**2. **Which of the following is true?

**a. **The pop operation on a stack is considerably simpler than the remove operation on a queue.

**b. **The contents of a queue can wrap around, while those of a stack cannot.

**c. **The top of a stack corresponds to the front of a queue.

**d. **In both the stack and the queue, items removed in sequence are taken from increasingly high index cells in the array.

**3. **What do LIFO and FIFO mean?

**4. **True or False: A stack or a queue often serves as the underlying mechanism on which an ADT array is based.

**5. **Assume an array is numbered with index 0 on the left. A queue representing a line of movie-goers, with the first to arrive numbered 1, has the ticket window on the right. Then

**a. **there is no numerical correspondence between the index numbers and the movie-goer numbers.

**b. **the array index numbers and the movie-goer numbers increase in opposite left-right directions.

**c. **the array index numbers correspond numerically to the locations in the line of movie-goers.

**d. **the movie-goers and the items in the array move in the same direction.

**6. **As other items are inserted and removed, does a particular item in a queue move along the array from lower to higher indices, or higher to lower?

**7. **Suppose you insert 15, 25, 35, and 45 into a queue. Then you remove three items. Which one is left?

**8. **True or False: Pushing and popping items on a stack and inserting and removing items in a queue all take O(N) time.

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175

**9. **A queue might be used to hold

**a. **the items to be sorted in an insertion sort.

**b. **reports of a variety of imminent attacks on the star ship Enterprise.

**c. **keystrokes made by a computer user writing a letter.

**d. **symbols in an algebraic expression being evaluated.

**10. **Inserting an item into a typical priority queue takes what big O time?

**11. **The term **a. **the highest priority items are inserted first.

**b. **the programmer must prioritize access to the underlying array.

**c. **the underlying array is sorted by the priority of the items.

**d. **the lowest priority items are deleted first.

**12. **True or False: At least one of the methods in the priorityQ.java program (Listing 4.6) uses a linear search.

**13. **One difference between a priority queue and an ordered array is that **a. **the lowest-priority item cannot be extracted easily from the array as it can from the priority queue.

**b. **the array must be ordered while the priority queue need not be.

**c. **the highest priority item can be extracted easily from the priority queue but not from the array.

**d. **All of the above.

**14. **Suppose you based a priority queue class on the OrdArray class in the orderedArray.java program (Listing 2.4) in Chapter 2, “Arrays.” This will buy you binary search capability. If you wanted the best performance for your priority queue, would you need to modify the OrdArray class?

**15. **A priority queue might be used to hold

**a. **passengers to be picked up by a taxi from different parts of the city.

**b. **keystrokes made at a computer keyboard.

**c. **squares on a chessboard in a game program.

**d. **planets in a solar system simulation.

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**Experiments**

Carrying out these experiments will help to provide insights into the topics covered in the chapter. No programming is involved.

**1. **Start with the initial configuration of the Queue Workshop applet. Alternately remove and insert items. (This way, you can reuse the deleted key value for the new item without typing it.) Notice how the group of four items crawls up to the top of the queue and then reappears at the bottom and keeps climbing.

**2. **Using the PriorityQ Workshop applet, figure out the positions of the Front and Rear arrows when the priority queue is full and when it is empty. Why can’t a priority queue wrap around like an ordinary queue?

**3. **Think about how you remember the events in your life. Are there times when they seem to be stored in your brain in a stack? In a queue? In a priority queue?

**Programming Projects**

Writing programs that solve the Programming Projects helps to solidify your understanding of the material and demonstrates how the chapter’s concepts are applied.

(As noted in the Introduction, qualified instructors may obtain completed solutions to the Programming Projects on the publisher’s Web site.) **4.1 **Write a method for the Queue class in the queue.java program (Listing 4.4) that displays the contents of the queue. Note that this does not mean simply displaying the contents of the underlying array. You should show the queue contents from the first item inserted to the last, without indicating to the viewer whether the sequence is broken by wrapping around the end of the array. Be careful that one item and no items display properly, no matter where front and rear are.

**4.2 **Create a Deque class based on the discussion of deques (double-ended queues) in this chapter. It should include insertLeft(), insertRight(), removeLeft(), removeRight(), isEmpty(), and isFull() methods. It will need to support wraparound at the end of the array, as queues do.

**4.3 **Write a program that implements a stack class that is based on the Deque class in Programming Project 4.2. This stack class should have the same methods and capabilities as the StackX class in the stack.java program (Listing 4.1).

**4.4 **The priority queue shown in Listing 4.6 features fast removal of the high-priority item but slow insertion of new items. Write a program with a revised PriorityQ class that has fast O(1) insertion time but slower removal of the high-priority item. Include a method that displays the contents of the priority queue, as suggested in Programming Project 4.1.

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177

**4.5 **Queues are often used to simulate the flow of people, cars, airplanes, transactions, and so on. Write a program that models checkout lines at a supermarket, using the Queue class from the queue.java program (Listing 4.4). Several lines of customers should be displayed; you can use the display() method of Programming Project 4.1. You can add a new customer by pressing a key. You’ll need to determine how the customer will decide which line to join. The checkers will take random amounts of time to process each customer (presumably depending on how many groceries the customer has). Once checked out, the customer is removed from the line. For simplicity, you can simulate the passing of time by pressing a key. Perhaps every keypress indicates the passage of one minute. (Java, of course, has more sophisticated ways to handle time.)

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**5**

**IN THIS CHAPTER**

• Links

Linked Lists

• A Simple Linked List

• Finding and Deleting

Specified Links

In Chapter 2, “Arrays,” we saw that arrays had certain

• Double-Ended Lists

disadvantages as data storage structures. In an unordered array, searching is slow, whereas in an ordered array, inser-

• Linked-List Efficiency

tion is slow. In both kinds of arrays, deletion is slow. Also, the size of an array can’t be changed after it’s created.

• Abstract Data Types

In this chapter we’ll look at a data storage structure that

• Sorted Lists

solves some of these problems: the

• Doubly Linked Lists

are probably the second most commonly used general-

purpose storage structures after arrays.

• Iterators

The linked list is a versatile mechanism suitable for use in many kinds of general-purpose databases. It can also replace an array as the basis for other storage structures such as stacks and queues. In fact, you can use a linked list in many cases in which you use an array, unless you need frequent random access to individual items using an index.

Linked lists aren’t the solution to all data storage problems, but they are surprisingly versatile and conceptually simpler than some other popular structures such as trees. We’ll investigate their strengths and weaknesses as we go along.

In this chapter we’ll look at simple linked lists, double-ended lists, sorted lists, doubly linked lists, and lists with iterators (an approach to random access to list elements).

We’ll also examine the idea of Abstract Data Types (ADTs), and see how stacks and queues can be viewed as ADTs and how they can be implemented as linked lists instead of arrays.

**Links**

In a linked list, each data item is embedded in a

Because there are many similar links in a list, it makes sense to use a separate class for them, distinct from the

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linked list itself. Each Link object contains a reference (usually called next) to the next link in the list. A field in the list itself contains a reference to the first link. This relationship is shown in Figure 5.1.

Linked List

Link

Link

Link

Link

Data

Data

Data

Data

first

next

next

next

next

Null

**FIGURE 5.1 **

Here’s part of the definition of a class Link. It contains some data and a reference to the next link:

class Link

{

public int iData; // data

public double dData; // data

public Link next; // reference to next link

}

This kind of class definition is sometimes called

We show only two data items in the link: an int and a double. In a typical application there would be many more. A personnel record, for example, might have name, address, Social Security number, title, salary, and many other fields. Often an object of a class that contains this data is used instead of the items: class Link

{

public inventoryItem iI; // object holding data

public Link next; // reference to next link

}

**References and Basic Types**

You can easily get confused about references in the context of linked lists, so let’s review how they work.

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Links

181

Being able to put a field of type Link inside the class definition of this same type may seem odd. Wouldn’t the compiler be confused? How can it figure out how big to make a Link object if a link contains a link and the compiler doesn’t already know how big a Link object is?

The answer is that in Java a Link object doesn’t really contain another Link object, although it may look like it does. The next field of type Link is only a

A reference is a number that

Note that in Java, primitive types such as int and double are stored quite differently than objects. Fields containing primitive types do not contain references, but actual numerical values like 7 or 3.14159. A variable definition like double salary = 65000.00;

creates a space in memory and puts the number 65000.00 into this space. However, a reference to an object like

Link aLink = someLink;

puts a reference to an object of type Link, called someLink, into aLink. The someLink object itself is located elsewhere. It isn’t moved, or even created, by this statement; it must have been created before. To create an object, you must always use new: Link someLink = new Link();

Even the someLink field doesn’t hold an object; it’s still just a reference. The object is somewhere else in memory, as shown in Figure 5.2.

Other languages, such as C++, handle objects quite differently than Java. In C++ a field like

Link next;

actually contains an object of type Link. You can’t write a self-referential class definition in C++ (although you can put a pointer to a Link in class Link; a pointer is similar to a reference). C++ programmers should keep in mind how Java handles objects; this usage may be counter-intuitive.

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aLink

someLink

aLink and

someLink

refer to

object of type Link

an object

of type

link

Memory

**FIGURE 5.2 **

**Relationship, Not Position**

Let’s examine one of the major ways in which linked lists differ from arrays. In an array each item occupies a particular position. This position can be directly accessed using an index number. It’s like a row of houses: You can find a particular house using its address.

In a list the only way to find a particular element is to follow along the chain of elements. It’s more like human relations. Maybe you ask Harry where Bob is. Harry doesn’t know, but he thinks Jane might know, so you go and ask Jane. Jane saw Bob leave the office with Sally, so you call Sally’s cell phone. She dropped Bob off at

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The LinkList Workshop Applet

183

Peter’s office, so…but you get the idea. You can’t access a data item directly; you must use relationships between the items to locate it. You start with the first item, go to the second, then the third, until you find what you’re looking for.

**The LinkList Workshop Applet**

The LinkList Workshop applet provides three list operations. You can insert a new data item, search for a data item with a specified key, and delete a data item with a specified key. These operations are the same ones we explored in the Array Workshop applet in Chapter 2; they’re suitable for a general-purpose database application.

Figure 5.3 shows how the LinkList Workshop applet looks when it’s started. Initially, there are 13 links on the list.

**FIGURE 5.3 **

**The Insert Button**

If you think 13 is an unlucky number, you can insert a new link. Press the Ins button, and you’ll be prompted to enter a key value between 0 and 999. Subsequent presses will generate a link with this data in it, as shown in Figure 5.4.

In this version of a linked list, new links are always inserted at the beginning of the list. This is the simplest approach, although you can also insert links anywhere in the list, as we’ll see later.

A final press on Ins will redraw the list so the newly inserted link lines up with the other links. This redrawing doesn’t represent anything happening in the program itself, it just makes the display neater.

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**FIGURE 5.4 **

**The Find Button**

The Find button allows you to find a link with a specified key value. When prompted, type in the value of an existing link, preferably one somewhere in the middle of the list. As you continue to press the button, you’ll see the red arrow move along the list, looking for the link. A message informs you when the arrow finds the link. If you type a non-existent key value, the arrow will search all the way to the end of the list before reporting that the item can’t be found.

**The Delete Button**

You can also delete a key with a specified value. Type in the value of an existing link and repeatedly press Del. Again, the arrow will move along the list, looking for the link. When the arrow finds the link, it simply removes that link and connects the arrow from the previous link straight across to the following link. This is how links are removed: The reference to the preceding link is changed to point to the following link.

A final keypress redraws the picture, but again redrawing just provides evenly spaced links for aesthetic reasons; the length of the arrows doesn’t correspond to anything in the program.

**NOTE**

The LinkList Workshop applet can create both unsorted and sorted lists. Unsorted is the default. We’ll show how to use the applet for sorted lists when we discuss them later in this chapter.

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A Simple Linked List

185

**A Simple Linked List**

Our first example program, linkList.java, demonstrates a simple linked list. The only operations allowed in this version of a list are

• Inserting an item at the beginning of the list

• Deleting the item at the beginning of the list

• Iterating through the list to display its contents These operations are fairly easy to carry out, so we’ll start with them. (As we’ll see later, these operations are also all you need to use a linked list as the basis for a stack.)

Before we get to the complete linkList.java program, we’ll look at some important parts of the Link and LinkList classes.

**The **Link **Class**

You’ve already seen the data part of the Link class. Here’s the complete class definition:

class Link

{

public int iData; // data item

public double dData; // data item

public Link next; // next link in list

// -------------------------------------------------------------

public Link(int id, double dd) // constructor

{

iData = id; // initialize data

dData = dd; // (‘next’ is automatically

} // set to null)

// -------------------------------------------------------------

public void displayLink() // display ourself

{

System.out.print(“{“ + iData + “, “ + dData + “} “);

}

} // end class Link

In addition to the data, there’s a constructor and a method, displayLink(), that displays the link’s data in the format {22, 33.9}. Object purists would probably object to naming this method displayLink(), arguing that it should be simply display(). Using the shorter name would be in the spirit of polymorphism, but it makes the listing somewhat harder to understand when you see a statement like

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current.display();

and you’ve forgotten whether current is a Link object, a LinkList object, or something else.

The constructor initializes the data. There’s no need to initialize the next field because it’s automatically set to null when it’s created. (However, you could set it to null explicitly, for clarity.) The null value means it doesn’t refer to anything, which is the situation until the link is connected to other links.

We’ve made the storage type of the Link fields (iData and so on) public. If they were private, we would need to provide public methods to access them, which would require extra code, thus making the listing longer and harder to read. Ideally, for security we would probably want to restrict Link-object access to methods of the LinkList class. However, without an inheritance relationship between these classes, that’s not very convenient. We could use the default access specifier (no keyword) to give the data

**The **LinkList **Class**

The LinkList class contains only one data item: a reference to the first link on the list. This reference is called first. It’s the only permanent information the list maintains about the location of any of the links. It finds the other links by following the chain of references from first, using each link’s next field: class LinkList

{

private Link first; // ref to first link on list

// -------------------------------------------------------------

public void LinkList() // constructor

{

first = null; // no items on list yet

}

// -------------------------------------------------------------

public boolean isEmpty() // true if list is empty

{

return (first==null);

}

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187

// -------------------------------------------------------------

// ... other methods go here

}

The constructor for LinkList sets first to null. This isn’t really necessary because, as we noted, references are set to null automatically when they’re created. However, the explicit constructor makes it clear that this is how first begins.

When first has the value null, we know there are no items on the list. If there were any items, first would contain a reference to the first one. The isEmpty() method uses this fact to determine whether the list is empty.

**The **insertFirst() **Method**

The insertFirst() method of LinkList inserts a new link at the beginning of the list.

This is the easiest place to insert a link because first already points to the first link.

To insert the new link, we need only set the next field in the newly created link to point to the old first link and then change first so it points to the newly created link. This situation is shown in Figure 5.5.

42

42

7

7

98

98

14

14

first

next

next

next

next

Null

a) Before Insertion

42

7

98

14

first

next

next

next

next

Null

❶

Link

❷

33

next

b) After Insertion

**FIGURE 5.5 **

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In insertFirst() we begin by creating the new link using the data passed as arguments. Then we change the link references as we just noted:

// insert at start of list

public void insertFirst(int id, double dd)

{ // make new link

Link newLink = new Link(id, dd);

newLink.next = first; // newLink --> old first first = newLink; // first --> newLink

}

The --> arrows in the comments in the last two statements mean that a link (or the first field) connects to the next (downstream) link. (In doubly linked lists we’ll see upstream connections as well, symbolized by <-- arrows.) Compare these two statements with Figure 5.5. Make sure you understand how the statements cause the links to be changed, as shown in the figure. This kind of reference manipulation is the heart of linked-list algorithms.

**The **deleteFirst() **Method**

The deleteFirst() method is the reverse of insertFirst(). It disconnects the first link by rerouting first to point to the second link. This second link is found by looking at the next field in the first link:

public Link deleteFirst() // delete first item

{ // (assumes list not empty) Link temp = first; // save reference to link first = first.next; // delete it: first-->old next return temp; // return deleted link

}

The second statement is all you need to remove the first link from the list. We choose to also return the link, for the convenience of the user of the linked list, so we save it in temp before deleting it and return the value of temp. Figure 5.6 shows how first is rerouted to delete the object.

In C++ and similar languages, you would need to worry about deleting the link itself after it was disconnected from the list. It’s in memory somewhere, but now nothing refers to it. What will become of it? In Java, the garbage collection process will destroy it at some point in the future; it’s not your responsibility.

Notice that the deleteFirst() method assumes the list is not empty. Before calling it, your program should verify this fact with the isEmpty() method.

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A Simple Linked List

189

27

94

6

33

first

Null

a) Before Deletion

94

6

33

first

Null

b) After Deletion

**FIGURE 5.6 **

**The **displayList() **Method**

To display the list, you start at first and follow the chain of references from link to link. A variable current points to (or technically

changes current to point to the next link because that’s what’s in the next field in each link. Here’s the entire displayList() method:

public void displayList()

{

System.out.print(“List (first-->last): “);

Link current = first; // start at beginning of list while(current != null) // until end of list,

{

current.displayLink(); // print data

current = current.next; // move to next link

}

System.out.println(“”);

}

The end of the list is indicated by the next field in the last link pointing to null rather than another link. How did this field get to be null? It started that way when the link was created and was never given any other value because it was always at

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the end of the list. The while loop uses this condition to terminate itself when it reaches the end of the list. Figure 5.7 shows how current steps along the list.

first

next

next

next

next

Null

a)Before current = current.next

current

first

next

next

next

next

Null

b)After current = current.next

current

**FIGURE 5.7 **

At each link, the displayList() method calls the displayLink() method to display the data in the link.

**The **linkList.java **Program**

Listing 5.1 shows the complete linkList.java program. You’ve already seen all the components except the main() routine.

**LISTING 5.1**

The linkList.java Program

// linkList.java

// demonstrates linked list

// to run this program: C>java LinkListApp

////////////////////////////////////////////////////////////////

class Link

{

public int iData; // data item (key)

public double dData; // data item

public Link next; // next link in list

// -------------------------------------------------------------

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A Simple Linked List

191

**LISTING 5.1**

Continued

public Link(int id, double dd) // constructor

{

iData = id; // initialize data

dData = dd; // (‘next’ is automatically

} // set to null)

// -------------------------------------------------------------

public void displayLink() // display ourself

{

System.out.print(“{“ + iData + “, “ + dData + “} “);

}

} // end class Link

////////////////////////////////////////////////////////////////

class LinkList

{

private Link first; // ref to first link on list

// -------------------------------------------------------------

public LinkList() // constructor

{

first = null; // no items on list yet

}

// -------------------------------------------------------------

public boolean isEmpty() // true if list is empty

{

return (first==null);

}

// -------------------------------------------------------------

// insert at start of list

public void insertFirst(int id, double dd)

{ // make new link

Link newLink = new Link(id, dd);

newLink.next = first; // newLink --> old first first = newLink; // first --> newLink

}

// -------------------------------------------------------------

public Link deleteFirst() // delete first item

{ // (assumes list not empty) Link temp = first; // save reference to link first = first.next; // delete it: first-->old next return temp; // return deleted link

}

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**LISTING 5.1**

Continued

// -------------------------------------------------------------

public void displayList()

{

System.out.print(“List (first-->last): “);

Link current = first; // start at beginning of list while(current != null) // until end of list,

{

current.displayLink(); // print data

current = current.next; // move to next link

}

System.out.println(“”);

}

// -------------------------------------------------------------

} // end class LinkList

////////////////////////////////////////////////////////////////

class LinkListApp

{

public static void main(String[] args)

{

LinkList theList = new LinkList(); // make new list theList.insertFirst(22, 2.99); // insert four items theList.insertFirst(44, 4.99);

theList.insertFirst(66, 6.99);

theList.insertFirst(88, 8.99);

theList.displayList(); // display list while( !theList.isEmpty() ) // until it’s empty,

{

Link aLink = theList.deleteFirst(); // delete link System.out.print(“Deleted “); // display it aLink.displayLink();

System.out.println(“”);

}

theList.displayList(); // display list

} // end main()

} // end class LinkListApp

////////////////////////////////////////////////////////////////

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193

In main() we create a new list, insert four new links into it with insertFirst(), and display it. Then, in the while loop, we remove the items one by one with deleteFirst() until the list is empty. The empty list is then displayed. Here’s the output from linkList.java:

List (first-->last): {88, 8.99} {66, 6.99} {44, 4.99} {22, 2.99}

Deleted {88, 8.99}

Deleted {66, 6.99}

Deleted {44, 4.99}

Deleted {22, 2.99}

List (first-->last):

**Finding and Deleting Specified Links**

Our next example program adds methods to search a linked list for a data item with a specified key value and to delete an item with a specified key value. These, along with insertion at the start of the list, are the same operations carried out by the LinkList Workshop applet. The complete linkList2.java program is shown in Listing 5.2.

**LISTING 5.2**

The linkList2.java Program

// linkList2.java

// demonstrates linked list

// to run this program: C>java LinkList2App

////////////////////////////////////////////////////////////////

class Link

{

public int iData; // data item (key)

public double dData; // data item

public Link next; // next link in list

// -------------------------------------------------------------

public Link(int id, double dd) // constructor

{

iData = id;

dData = dd;

}

// -------------------------------------------------------------

public void displayLink() // display ourself

{

System.out.print(“{“ + iData + “, “ + dData + “} “);

}

} // end class Link

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**LISTING 5.2**

Continued

////////////////////////////////////////////////////////////////

class LinkList

{

private Link first; // ref to first link on list

// -------------------------------------------------------------

public LinkList() // constructor

{

first = null; // no links on list yet

}

// -------------------------------------------------------------

public void insertFirst(int id, double dd)

{ // make new link

Link newLink = new Link(id, dd);

newLink.next = first; // it points to old first link first = newLink; // now first points to this

}

// -------------------------------------------------------------

public Link find(int key) // find link with given key

{ // (assumes non-empty list) Link current = first; // start at ‘first’

while(current.iData != key) // while no match,

{

if(current.next == null) // if end of list,

return null; // didn’t find it

else // not end of list, current = current.next; // go to next link

}

return current; // found it

}

// -------------------------------------------------------------

public Link delete(int key) // delete link with given key

{ // (assumes non-empty list) Link current = first; // search for link Link previous = first;

while(current.iData != key)

{

if(current.next == null)

return null; // didn’t find it

else

{

previous = current; // go to next link

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**LISTING 5.2**

Continued

current = current.next;

}

} // found it

if(current == first) // if first link, first = first.next; // change first

else // otherwise,

previous.next = current.next; // bypass it

return current;

}

// -------------------------------------------------------------

public void displayList() // display the list

{

System.out.print(“List (first-->last): “);

Link current = first; // start at beginning of list while(current != null) // until end of list,

{

current.displayLink(); // print data

current = current.next; // move to next link

}

System.out.println(“”);

}

// -------------------------------------------------------------

} // end class LinkList

////////////////////////////////////////////////////////////////

class LinkList2App

{

public static void main(String[] args)

{

LinkList theList = new LinkList(); // make list

theList.insertFirst(22, 2.99); // insert 4 items theList.insertFirst(44, 4.99);

theList.insertFirst(66, 6.99);

theList.insertFirst(88, 8.99);

theList.displayList(); // display list Link f = theList.find(44); // find item

if( f != null)

System.out.println(“Found link with key “ + f.iData); else

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**LISTING 5.2**

Continued

System.out.println(“Can’t find link”);

Link d = theList.delete(66); // delete item

if( d != null )

System.out.println(“Deleted link with key “ + d.iData); else

System.out.println(“Can’t delete link”);

theList.displayList(); // display list

} // end main()

} // end class LinkList2App

////////////////////////////////////////////////////////////////

The main() routine makes a list, inserts four items, and displays the resulting list. It then searches for the item with key 44, deletes the item with key 66, and displays the list again. Here’s the output:

List (first-->last): {88, 8.99} {66, 6.99} {44, 4.99} {22, 2.99}

Found link with key 44

Deleted link with key 66

List (first-->last): {88, 8.99} {44, 4.99} {22, 2.99}

**The **find() **Method**

The find() method works much like the displayList() method in the linkList.java program. The reference current initially points to first and then steps its way along the links by setting itself repeatedly to current.next. At each link, find() checks whether that link’s key is the one it’s looking for. If the key is found, it returns with a reference to that link. If find() reaches the end of the list without finding the desired link, it returns null.

**The **delete() **Method**

The delete() method is similar to find() in the way it searches for the link to be deleted. However, it needs to maintain a reference not only to the current link (current), but to the link preceding the current link (previous). It does so because, if it deletes the current link, it must connect the preceding link to the following link, as shown in Figure 5.8. The only way to tell where the preceding link is located is to maintain a reference to it.

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17

17

44

44

98

98

73

73

First

Next

Next

Next

Next

Null

a) Before deletion

Previous

Current

17

17

44

44

73

73

First

Next

Next

Next

Null

b) After deletion

Previous

Current

**FIGURE 5.8 **

At each cycle through the while loop, just before current is set to current.next, previous is set to current. This keeps it pointing at the link preceding current.

To delete the current link once it’s found, the next field of the previous link is set to the next link. A special case arises if the current link is the first link because the first link is pointed to by the LinkList’s first field and not by another link. In this case the link is deleted by changing first to point to first.next, as we saw in the linkList.java program with the deleteFirst() method. Here’s the code that covers these two possibilities:

// found it

if(current == first) // if first link, first = first.next; // change first

else // otherwise,

previous.next = current.next; // bypass link

**Other Methods**

We’ve seen methods to insert and delete items at the start of a list, and to find a specified item and delete a specified item. You can imagine other useful list methods.

For example, an insertAfter() method could find a link with a specified key value and insert a new link following it. We’ll see such a method when we talk about list iterators at the end of this chapter.

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**Double-Ended Lists**

A double-ended list is similar to an ordinary linked list, but it has one additional feature: a reference to the last link as well as to the first. Figure 5.9 shows such a list.

next

next

next

next

first

Null

last

**FIGURE 5.9 **

The reference to the last link permits you to insert a new link directly at the end of the list as well as at the beginning. Of course, you can insert a new link at the end of an ordinary single-ended list by iterating through the entire list until you reach the end, but this approach is inefficient.

Access to the end of the list as well as the beginning makes the double-ended list suitable for certain situations that a single-ended list can’t handle efficiently. One such situation is implementing a queue; we’ll see how this technique works in the next section.

Listing 5.3 contains the firstLastList.java program, which demonstrates a double-ended list. (Incidentally, don’t confuse the double-ended list with the doubly linked list, which we’ll explore later in this chapter.)

**LISTING 5.3**

The firstLastList.java Program

// firstLastList.java

// demonstrates list with first and last references

// to run this program: C>java FirstLastApp

////////////////////////////////////////////////////////////////

class Link

{

public long dData; // data item

public Link next; // next link in list

// -------------------------------------------------------------

public Link(long d) // constructor

{ dData = d; }

// -------------------------------------------------------------

public void displayLink() // display this link

{ System.out.print(dData + “ “); }

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Double-Ended Lists

199

**LISTING 5.3**

Continued

// -------------------------------------------------------------

} // end class Link

////////////////////////////////////////////////////////////////

class FirstLastList

{

private Link first; // ref to first link private Link last; // ref to last link

// -------------------------------------------------------------

public FirstLastList() // constructor

{

first = null; // no links on list yet last = null;

}

// -------------------------------------------------------------

public boolean isEmpty() // true if no links

{ return first==null; }

// -------------------------------------------------------------

public void insertFirst(long dd) // insert at front of list

{

Link newLink = new Link(dd); // make new link

if( isEmpty() ) // if empty list,

last = newLink; // newLink <-- last

newLink.next = first; // newLink --> old first first = newLink; // first --> newLink

}

// -------------------------------------------------------------

public void insertLast(long dd) // insert at end of list

{

Link newLink = new Link(dd); // make new link

if( isEmpty() ) // if empty list,

first = newLink; // first --> newLink else

last.next = newLink; // old last --> newLink last = newLink; // newLink <-- last

}

// -------------------------------------------------------------

public long deleteFirst() // delete first link

{ // (assumes non-empty list) long temp = first.dData;

if(first.next == null) // if only one item

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**LISTING 5.3**

Continued

last = null; // null <-- last

first = first.next; // first --> old next return temp;

}

// -------------------------------------------------------------

public void displayList()

{

System.out.print(“List (first-->last): “);

Link current = first; // start at beginning while(current != null) // until end of list,

{

current.displayLink(); // print data

current = current.next; // move to next link

}

System.out.println(“”);

}

// -------------------------------------------------------------

} // end class FirstLastList

////////////////////////////////////////////////////////////////

class FirstLastApp

{

public static void main(String[] args)

{ // make a new list

FirstLastList theList = new FirstLastList();

theList.insertFirst(22); // insert at front

theList.insertFirst(44);

theList.insertFirst(66);

theList.insertLast(11); // insert at rear

theList.insertLast(33);

theList.insertLast(55);

theList.displayList(); // display the list

theList.deleteFirst(); // delete first two items theList.deleteFirst();

theList.displayList(); // display again

} // end main()

} // end class FirstLastApp

////////////////////////////////////////////////////////////////

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201

For simplicity, in this program we’ve reduced the number of data items in each link from two to one. This makes it easier to display the link contents. (Remember that in a serious program there would be many more data items, or a reference to another object containing many data items.)

This program inserts three items at the front of the list, inserts three more at the end, and displays the resulting list. It then deletes the first two items and displays the list again. Here’s the output:

List (first-->last): 66 44 22 11 33 55

List (first-->last): 22 11 33 55

Notice how repeated insertions at the front of the list reverse the order of the items, while repeated insertions at the end preserve the order.

The double-ended list class is called the FirstLastList. As discussed, it has two data items, first and last, which point to the first item and the last item in the list. If there is only one item in the list, both first and last point to it, and if there are no items, they are both null.

The class has a new method, insertLast(), that inserts a new item at the end of the list. This process involves modifying last.next to point to the new link and then changing last to point to the new link, as shown in Figure 5.10.

next

next

next

next

first

Null

last

a) Before insertion

next

next

next

next

first

First

last

❶

b) After insertion

❷

Null

**FIGURE 5.10 **

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The insertion and deletion routines are similar to those in a single-ended list.

However, both insertion routines must watch out for the special case when the list is empty prior to the insertion. That is, if isEmpty() is true, then insertFirst() must set last to the new link, and insertLast() must set first to the new link.

If inserting at the beginning with insertFirst(), first is set to point to the new link, although when inserting at the end with insertLast(), last is set to point to the new link. Deleting from the start of the list is also a special case if it’s the last item on the list: last must be set to point to null in this case.

Unfortunately, making a list double-ended doesn’t help you to delete the last link because there is still no reference to the next-to-last link, whose next field would need to be changed to null if the last link were deleted. To conveniently delete the last link, you would need a doubly linked list, which we’ll look at soon. (Of course, you could also traverse the entire list to find the last link, but that’s not very efficient.)

**Linked-List Efficiency**

Insertion and deletion at the beginning of a linked list are very fast. They involve changing only one or two references, which takes O(1) time.

Finding, deleting, or inserting next to a specific item requires searching through, on the average, half the items in the list. This requires O(N) comparisons. An array is also O(N) for these operations, but the linked list is nevertheless faster because nothing needs to be moved when an item is inserted or deleted. The increased efficiency can be significant, especially if a copy takes much longer than a comparison.

Of course, another important advantage of linked lists over arrays is that a linked list uses exactly as much memory as it needs and can expand to fill all of available memory. The size of an array is fixed when it’s created; this usually leads to inefficiency because the array is too large, or to running out of room because the array is too small. Vectors, which are expandable arrays, may solve this problem to some extent, but they usually expand in fixed-sized increments (such as doubling the size of the array whenever it’s about to overflow). This solution is still not as efficient a use of memory as a linked list.

**Abstract Data Types**

In this section we’ll shift gears and discuss a topic that’s more general than linked lists: Abstract Data Types (ADTs). What is an ADT? Roughly speaking, it’s a way of looking at a data structure: focusing on what it does and ignoring how it does its job.

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Stacks and queues are examples of ADTs. We’ve already seen that both stacks and queues can be implemented using arrays. Before we return to a discussion of ADTs, let’s see how stacks and queues can be implemented using linked lists. This discussion will demonstrate the “abstract” nature of stacks and queues: how they can be considered separately from their implementation.

**A Stack Implemented by a Linked List**

When we created a stack in Chapter 4, “Stacks and Queues,” we used an ordinary Java array to hold the stack’s data. The stack’s push() and pop() operations were actually carried out by array operations such as

arr[++top] = data;

and

data = arr[top--];

which insert data into, and take it out of, an array.

We can also use a linked list to hold a stack’s data. In this case the push() and pop() operations would be carried out by operations like

theList.insertFirst(data)

and

data = theList.deleteFirst()

The user of the stack class calls push() and pop() to insert and delete items without knowing, or needing to know, whether the stack is implemented as an array or as a linked list. Listing 5.4 shows how a stack class called LinkStack can be implemented using the LinkList class instead of an array. (Object purists would argue that the name LinkStack should be simply Stack because users of this class shouldn’t need to know that it’s implemented as a list.)

**LISTING 5.4**

The linkStack.java Program

// linkStack.java

// demonstrates a stack implemented as a list

// to run this program: C>java LinkStackApp

////////////////////////////////////////////////////////////////

class Link

{

public long dData; // data item

public Link next; // next link in list

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**LISTING 5.4**

Continued

// -------------------------------------------------------------

public Link(long dd) // constructor

{ dData = dd; }

// -------------------------------------------------------------

public void displayLink() // display ourself

{ System.out.print(dData + “ “); }

} // end class Link

////////////////////////////////////////////////////////////////

class LinkList

{

private Link first; // ref to first item on list

// -------------------------------------------------------------

public LinkList() // constructor

{ first = null; } // no items on list yet

// -------------------------------------------------------------

public boolean isEmpty() // true if list is empty

{ return (first==null); }

// -------------------------------------------------------------

public void insertFirst(long dd) // insert at start of list

{ // make new link

Link newLink = new Link(dd);

newLink.next = first; // newLink --> old first first = newLink; // first --> newLink

}

// -------------------------------------------------------------

public long deleteFirst() // delete first item

{ // (assumes list not empty) Link temp = first; // save reference to link first = first.next; // delete it: first-->old next return temp.dData; // return deleted link

}

// -------------------------------------------------------------

public void displayList()

{

Link current = first; // start at beginning of list while(current != null) // until end of list,

{

current.displayLink(); // print data

current = current.next; // move to next link

}

System.out.println(“”);

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**LISTING 5.4**

Continued

}

// -------------------------------------------------------------

} // end class LinkList

////////////////////////////////////////////////////////////////

class LinkStack

{

private LinkList theList;

//--------------------------------------------------------------

public LinkStack() // constructor

{

theList = new LinkList();

}

//--------------------------------------------------------------

public void push(long j) // put item on top of stack

{

theList.insertFirst(j);

}

//--------------------------------------------------------------

public long pop() // take item from top of stack

{

return theList.deleteFirst();

}

//--------------------------------------------------------------

public boolean isEmpty() // true if stack is empty

{

return ( theList.isEmpty() );

}

//--------------------------------------------------------------

public void displayStack()

{

System.out.print(“Stack (top-->bottom): “);

theList.displayList();

}

//--------------------------------------------------------------

} // end class LinkStack

////////////////////////////////////////////////////////////////

class LinkStackApp

{

public static void main(String[] args)

{

LinkStack theStack = new LinkStack(); // make stack

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**LISTING 5.4**

Continued

theStack.push(20); // push items theStack.push(40);

theStack.displayStack(); // display stack theStack.push(60); // push items theStack.push(80);

theStack.displayStack(); // display stack theStack.pop(); // pop items

theStack.pop();

theStack.displayStack(); // display stack

} // end main()

} // end class LinkStackApp

////////////////////////////////////////////////////////////////

The main() routine creates a stack object, pushes two items on it, displays the stack, pushes two more items, and displays the stack again. Finally, it pops two items and displays the stack a third time. Here’s the output: Stack (top-->bottom): 40 20

Stack (top-->bottom): 80 60 40 20

Stack (top-->bottom): 40 20

Notice the overall organization of this program. The main() routine in the LinkStackApp class relates only to the LinkStack class. The LinkStack class relates only to the LinkList class. There’s no communication between main() and the LinkList class.

More specifically, when a statement in main() calls the push() operation in the LinkStack class, this method in turn calls insertFirst() in the LinkList class to actually insert data. Similarly, pop() calls deleteFirst() to delete an item, and displayStack() calls displayList() to display the stack. To the class user, writing code in main(), there is no difference between using the list-based LinkStack class and using the array-based stack class from the stack.java program (Listing 4.1) in Chapter 4.

**A Queue Implemented by a Linked List**

Here’s a similar example of an ADT implemented with a linked list. Listing 5.5 shows a queue implemented as a double-ended linked list.

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**LISTING 5.5**

The linkQueue.java Program

// linkQueue.java

// demonstrates queue implemented as double-ended list

// to run this program: C>java LinkQueueApp

////////////////////////////////////////////////////////////////

class Link

{

public long dData; // data item

public Link next; // next link in list

// -------------------------------------------------------------

public Link(long d) // constructor

{ dData = d; }

// -------------------------------------------------------------

public void displayLink() // display this link

{ System.out.print(dData + “ “); }

// -------------------------------------------------------------

} // end class Link

////////////////////////////////////////////////////////////////

class FirstLastList

{

private Link first; // ref to first item private Link last; // ref to last item

// -------------------------------------------------------------

public FirstLastList() // constructor

{

first = null; // no items on list yet last = null;

}

// -------------------------------------------------------------

public boolean isEmpty() // true if no links

{ return first==null; }

// -------------------------------------------------------------

public void insertLast(long dd) // insert at end of list

{

Link newLink = new Link(dd); // make new link

if( isEmpty() ) // if empty list,

first = newLink; // first --> newLink else

last.next = newLink; // old last --> newLink last = newLink; // newLink <-- last

}

// -------------------------------------------------------------

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**LISTING 5.5**

Continued

public long deleteFirst() // delete first link

{ // (assumes non-empty list) long temp = first.dData;

if(first.next == null) // if only one item

last = null; // null <-- last

first = first.next; // first --> old next return temp;

}

// -------------------------------------------------------------

public void displayList()

{

Link current = first; // start at beginning while(current != null) // until end of list,

{

current.displayLink(); // print data

current = current.next; // move to next link

}

System.out.println(“”);

}

// -------------------------------------------------------------

} // end class FirstLastList

////////////////////////////////////////////////////////////////

class LinkQueue

{

private FirstLastList theList;

//--------------------------------------------------------------

public LinkQueue() // constructor

{ theList = new FirstLastList(); } // make a 2-ended list

//--------------------------------------------------------------

public boolean isEmpty() // true if queue is empty

{ return theList.isEmpty(); }

//--------------------------------------------------------------

public void insert(long j) // insert, rear of queue

{ theList.insertLast(j); }

//--------------------------------------------------------------

public long remove() // remove, front of queue

{ return theList.deleteFirst(); }

//--------------------------------------------------------------

public void displayQueue()

{

System.out.print(“Queue (front-->rear): “);

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**LISTING 5.5**

Continued

theList.displayList();

}

//--------------------------------------------------------------

} // end class LinkQueue

////////////////////////////////////////////////////////////////

class LinkQueueApp

{

public static void main(String[] args)

{

LinkQueue theQueue = new LinkQueue();

theQueue.insert(20); // insert items theQueue.insert(40);

theQueue.displayQueue(); // display queue theQueue.insert(60); // insert items theQueue.insert(80);

theQueue.displayQueue(); // display queue theQueue.remove(); // remove items theQueue.remove();

theQueue.displayQueue(); // display queue

} // end main()

////////////////////////////////////////////////////////////////

The program creates a queue, inserts two items, inserts two more items, and removes two items; following each of these operations the queue is displayed. Here’s the output:

Queue (front-->rear): 20 40

Queue (front-->rear): 20 40 60 80

Queue (front-->rear): 60 80

Here the methods insert() and remove() in the LinkQueue class are implemented by the insertLast() and deleteFirst() methods of the FirstLastList class. We’ve substituted a linked list for the array used to implement the queue in the queue.java program (Listing 4.4) of Chapter 4.

The linkStack.java and linkQueue.java programs emphasize that stacks and queues are conceptual entities, separate from their implementations. A stack can be implemented equally well by an array or by a linked list. What’s important about a stack is

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the push() and pop() operations and how they’re used; it’s not the underlying mechanism used to implement these operations.

When would you use a linked list as opposed to an array as the implementation of a stack or queue? One consideration is how accurately you can predict the amount of data the stack or queue will need to hold. If this isn’t clear, the linked list gives you more flexibility than an array. Both are fast, so speed is probably not a major consideration.

**Data Types and Abstraction**

Where does the term

**Data Types**

The phrase

When you talk about a primitive type, you’re actually referring to two things: a data item with certain characteristics and permissible operations on that data. For example, type int variables in Java can have whole-number values between

–2,147,483,648 and +2,147,483,647, and the operators +, –, *, /, and so on can be applied to them. The data type’s permissible operations are an inseparable part of its identity; understanding the type means understanding what operations can be performed on it.

With the advent of object-oriented programming, you could now create your own data types using classes. Some of these data types represent numerical quantities that are used in ways similar to primitive types. You can, for example, define a class for time (with fields for hours, minutes, seconds), a class for fractions (with numerator and denominator fields), and a class for extra-long numbers (characters in a string represent the digits). All these classes can be added and subtracted like int and double, except that in Java you must use methods with functional notation like add() and sub() rather than operators like + and –.

The phrase

However, it is also applied to classes that don’t have this quantitative aspect. In fact,

By extension, when a data storage structure like a stack or queue is represented by a class, it too can be referred to as a data type. A stack is different in many ways from an int, but they are both defined as a certain arrangement of data and a set of operations on that data.

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**Abstraction**

The word

In object-oriented programming, then, an Abstract Data Type is a class considered without regard to its implementation. It’s a description of the data in the class (fields), a list of operations (methods) that can be carried out on that data, and instructions on how to use these operations. Specifically excluded are the details of how the methods carry out their tasks. As a class user, you’re told what methods to call, how to call them, and the results you can expect, but not how they work.

The meaning of

For the stack, the user knows that push() and pop() (and perhaps a few other methods) exist and how they work. The user doesn’t (at least not usually) need to know how push() and pop() work, or whether data is stored in an array, a linked list, or some other data structure like a tree.

**The Interface**

An ADT specification is often called an

**ADT Lists**

Now that we know what an Abstract Data Type is, we can mention another one: the

Don’t confuse the ADT list with the linked list we’ve been discussing in this chapter.

A list is defined by its interface: the specific methods used to interact with it. This interface can be implemented by various structures, including arrays and linked lists.

The list is an abstraction of such data structures.

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**ADTs as a Design Tool**

The ADT concept is a useful aid in the software design process. If you need to store data, start by considering the operations that need to be performed on that data. Do you need access to the last item inserted? The first one? An item with a specified key? An item in a certain position? Answering such questions leads to the definition of an ADT. Only after the ADT is completely defined should you worry about the details of how to represent the data and how to code the methods that access the data.

By decoupling the specification of the ADT from the implementation details, you can simplify the design process. You also make it easier to change the implementation at some future time. If a user relates only to the ADT interface, you should be able to change the implementation without “breaking” the user’s code.

Of course, once the ADT has been designed, the underlying data structure must be carefully chosen to make the specified operations as efficient as possible. If you need random access to element N, for example, the linked-list representation isn’t so good because random access isn’t an efficient operation for a linked list. You’d be better off with an array.

**NOTE**

Remember that the ADT concept is only a conceptual tool. Data storage structures are not divided cleanly into some that are ADTs and some that are used to implement ADTs. A linked list, for example, doesn’t need to be wrapped in a list interface to be useful; it can act as an ADT on its own, or it can be used to implement another data type such as a queue. A linked list can be implemented using an array, and an array-type structure can be implemented using a linked list. What’s an ADT and what’s a more basic structure must be determined in a given context.

**Sorted Lists**

In the linked lists we’ve seen thus far, there was no requirement that data be stored in order. However, for certain applications it’s useful to maintain the data in sorted order within the list. A list with this characteristic is called a

In a sorted list, the items are arranged in sorted order by key value. Deletion is often limited to the smallest (or the largest) item in the list, which is at the start of the list, although sometimes find() and delete() methods, which search through the list for specified links, are used as well.

In general you can use a sorted list in most situations in which you use a sorted array. The advantages of a sorted list over a sorted array are speed of insertion (because elements don’t need to be moved) and the fact that a list can expand to fill

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available memory, while an array is limited to a fixed size. However, a sorted list is somewhat more difficult to implement than a sorted array.

Later we’ll look at one application for sorted lists: sorting data. A sorted list can also be used to implement a priority queue, although a heap (see Chapter 12, “Heaps”) is a more common implementation.

The LinkList Workshop applet introduced at the beginning of this chapter demonstrates sorted as well as unsorted lists. To see how sorted lists work, use the New button to create a new list with about 20 links, and when prompted, click on the Sorted button. The result is a list with data in sorted order, as shown in Figure 5.11.

**FIGURE 5.11 **

Use the Ins button to insert a new item. Type in a value that will fall somewhere in the middle of the list. Watch as the algorithm traverses the links, looking for the appropriate insertion place. When it finds the correct location, it inserts the new link, as shown in Figure 5.12.

With the next press of Ins, the list will be redrawn to regularize its appearance. You can also find a specified link using the Find button and delete a specified link using the Del button.

**Java Code to Insert an Item in a Sorted List**

To insert an item in a sorted list, the algorithm must first search through the list until it finds the appropriate place to put the item: this is just before the first item that’s larger, as shown in Figure 5.12.

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**FIGURE 5.12 **

When the algorithm finds where to put it, the item can be inserted in the usual way by changing next in the new link to point to the next link and changing next in the previous link to point to the new link. However, we need to consider some special cases: The link might need to be inserted at the beginning of the list, or it might need to go at the end. Let’s look at the code:

public void insert(long key) // insert in order

{

Link newLink = new Link(key); // make new link

Link previous = null; // start at first

Link current = first;

// until end of list,

while(current != null && key > current.dData)

{ // or key > current, previous = current;

current = current.next; // go to next item

}

if(previous==null) // at beginning of list first = newLink; // first --> newLink else // not at beginning previous.next = newLink; // old prev --> newLink newLink.next = current; // newLink --> old current

} // end insert()

We need to maintain a previous reference as we move along, so we can modify the previous link’s next field to point to the new link. After creating the new link, we

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prepare to search for the insertion point by setting current to first in the usual way.

We also set previous to null; this step is important because later we’ll use this null value to determine whether we’re still at the beginning of the list.

The while loop is similar to those we’ve used before to search for the insertion point, but there’s an added condition. The loop terminates when the key of the link currently being examined (current.dData) is no longer smaller than the key of the link being inserted (key); this is the most usual case, where a key is inserted somewhere in the middle of the list.

However, the while loop also terminates if current is null. This happens at the end of the list (the next field of the last element is null), or if the list is empty to begin with (first is null).

When the while loop terminates, then, we may be at the beginning, the middle, or the end of the list, or the list may be empty.

If we’re at the beginning, or the list is empty, previous will be null; so we set first to the new link. Otherwise, we’re in the middle of the list, or at the end, and we set previous.next to the new link.

In any case we set the new link’s next field to current. If we’re at the end of the list, current is null, so the new link’s next field is appropriately set to this value.

**The **sortedList.java **Program**

The sortedList.java example shown in Listing 5.6 presents a SortedList class with insert(), remove(), and displayList() methods. Only the insert() routine is different from its counterpart in non-sorted lists.

**LISTING 5.6**

The sortedList.java Program

// sortedList.java

// demonstrates sorted list

// to run this program: C>java SortedListApp

////////////////////////////////////////////////////////////////

class Link

{

public long dData; // data item

public Link next; // next link in list

// -------------------------------------------------------------

public Link(long dd) // constructor

{ dData = dd; }

// -------------------------------------------------------------

public void displayLink() // display this link

{ System.out.print(dData + “ “); }

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**LISTING 5.6**

Continued

} // end class Link

////////////////////////////////////////////////////////////////

class SortedList

{

private Link first; // ref to first item on list

// -------------------------------------------------------------

public SortedList() // constructor

{ first = null; }

// -------------------------------------------------------------

public boolean isEmpty() // true if no links

{ return (first==null); }

// -------------------------------------------------------------

public void insert(long key) // insert, in order

{

Link newLink = new Link(key); // make new link

Link previous = null; // start at first

Link current = first;

// until end of list,

while(current != null && key > current.dData)

{ // or key > current, previous = current;

current = current.next; // go to next item

}

if(previous==null) // at beginning of list first = newLink; // first --> newLink else // not at beginning previous.next = newLink; // old prev --> newLink newLink.next = current; // newLink --> old current

} // end insert()

// -------------------------------------------------------------

public Link remove() // return & delete first link

{ // (assumes non-empty list) Link temp = first; // save first

first = first.next; // delete first

return temp; // return value

}

// -------------------------------------------------------------

public void displayList()

{

System.out.print(“List (first-->last): “);

Link current = first; // start at beginning of list

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**LISTING 5.6**

Continued

while(current != null) // until end of list,

{

current.displayLink(); // print data

current = current.next; // move to next link

}

System.out.println(“”);

}

} // end class SortedList

////////////////////////////////////////////////////////////////

class SortedListApp

{

public static void main(String[] args)

{ // create new list

SortedList theSortedList = new SortedList();

theSortedList.insert(20); // insert 2 items

theSortedList.insert(40);

theSortedList.displayList(); // display list

theSortedList.insert(10); // insert 3 more items theSortedList.insert(30);

theSortedList.insert(50);

theSortedList.displayList(); // display list

theSortedList.remove(); // remove an item

theSortedList.displayList(); // display list

} // end main()

} // end class SortedListApp

////////////////////////////////////////////////////////////////

In main() we insert two items with key values 20 and 40. Then we insert three more items, with values 10, 30, and 50. These values are inserted at the beginning of the list, in the middle, and at the end, showing that the insert() routine correctly handles these special cases. Finally, we remove one item, to show removal is always from the front of the list. After each change, the list is displayed. Here’s the output from sortedList.java:

List (first-->last): 20 40

List (first-->last): 10 20 30 40 50

List (first-->last): 20 30 40 50

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**Efficiency of Sorted Linked Lists**

Insertion and deletion of arbitrary items in the sorted linked list require O(N) comparisons (N/2 on the average) because the appropriate location must be found by stepping through the list. However, the minimum value can be found, or deleted, in O(1) time because it’s at the beginning of the list. If an application frequently accesses the minimum item, and fast insertion isn’t critical, then a sorted linked list is an effective choice. A priority queue might be implemented by a sorted linked list, for example.

**List Insertion Sort**

A sorted list can be used as a fairly efficient sorting mechanism. Suppose you have an array of unsorted data items. If you take the items from the array and insert them one by one into the sorted list, they’ll be placed in sorted order automatically. If you then remove them from the list and put them back in the array, the array will be sorted.

This type of sort turns out to be substantially more efficient than the more usual insertion sort within an array, described in Chapter 3, “Simple Sorting,” because fewer copies are necessary. It’s still an O(N2) process because inserting each item into the sorted list involves comparing a new item with an average of half the items already in the list, and there are N items to insert, resulting in about N2/4 comparisons. However, each item is copied only twice: once from the array to the list and once from the list to the array. N*2 copies compares favorably with the insertion sort within an array, where there are about N2 copies.

Listing 5.7 shows the listInsertionSort.java program, which starts with an array of unsorted items of type link, inserts them into a sorted list (using a constructor), and then removes them and places them back into the array.

**LISTING 5.7**

The listInsertionSort.java Program

// listInsertionSort.java

// demonstrates sorted list used for sorting

// to run this program: C>java ListInsertionSortApp

////////////////////////////////////////////////////////////////

class Link

{

public long dData; // data item

public Link next; // next link in list

// -------------------------------------------------------------

public Link(long dd) // constructor

{ dData = dd; }

// -------------------------------------------------------------

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**LISTING 5.7**

Continued

} // end class Link

////////////////////////////////////////////////////////////////

class SortedList

{

private Link first; // ref to first item on list

// -------------------------------------------------------------

public SortedList() // constructor (no args)

{ first = null; } // initialize list

// -------------------------------------------------------------

public SortedList(Link[] linkArr) // constructor (array

{ // as argument)

first = null; // initialize list for(int j=0; j<linkArr.length; j++) // copy array insert( linkArr[j] ); // to list

}

// -------------------------------------------------------------

public void insert(Link k) // insert (in order)

{

Link previous = null; // start at first

Link current = first;

// until end of list,

while(current != null && k.dData > current.dData)

{ // or key > current, previous = current;

current = current.next; // go to next item

}

if(previous==null) // at beginning of list first = k; // first --> k

else // not at beginning previous.next = k; // old prev --> k

k.next = current; // k --> old current

} // end insert()

// -------------------------------------------------------------

public Link remove() // return & delete first link

{ // (assumes non-empty list) Link temp = first; // save first

first = first.next; // delete first

return temp; // return value

}

// -------------------------------------------------------------

} // end class SortedList

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**LISTING 5.7**

Continued

////////////////////////////////////////////////////////////////

class ListInsertionSortApp

{

public static void main(String[] args)

{

int size = 10;

// create array of links

Link[] linkArray = new Link[size];

for(int j=0; j<size; j++) // fill array with links

{ // random number

int n = (int)(java.lang.Math.random()*99);

Link newLink = new Link(n); // make link

linkArray[j] = newLink; // put in array

}

// display array contents

System.out.print(“Unsorted array: “);

for(int j=0; j<size; j++)

System.out.print( linkArray[j].dData + “ “ );

System.out.println(“”);

// create new list

// initialized with array

SortedList theSortedList = new SortedList(linkArray); for(int j=0; j<size; j++) // links from list to array linkArray[j] = theSortedList.remove();

// display array contents

System.out.print(“Sorted Array: “);

for(int j=0; j<size; j++)

System.out.print(linkArray[j].dData + “ “);

System.out.println(“”);

} // end main()

} // end class ListInsertionSortApp

////////////////////////////////////////////////////////////////

This program displays the values in the array before the sorting operation and again afterward. Here’s some sample output:

Unsorted array: 59 69 41 56 84 15 86 81 37 35

Sorted array: 15 35 37 41 56 59 69 81 84 86

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The output will be different each time because the initial values are generated randomly.

A new constructor for SortedList takes an array of Link objects as an argument and inserts the entire contents of this array into the newly created list. By doing so, it helps make things easier for the client (the main() routine).

We’ve also made a change to the insert() routine in this program. It now accepts a Link object as an argument, rather than a long. We do this so we can store Link objects in the array and insert them directly into the list. In the sortedList.java program (Listing 5.6), it was more convenient to have the insert() routine create each Link object, using the long value passed as an argument.

The downside of the list insertion sort, compared with an array-based insertion sort, is that it takes somewhat more than twice as much memory: The array and linked list must be in memory at the same time. However, if you have a sorted linked list class handy, the list insertion sort is a convenient way to sort arrays that aren’t too large.

**Doubly Linked Lists**

Let’s examine another variation on the linked list: the

current=current.next

steps conveniently to the next link, but there’s no corresponding way to go to the previous link. Depending on the application, this limitation could pose problems.

For example, imagine a text editor in which a linked list is used to store the text.

Each text line on the screen is stored as a String object embedded in a link. When the editor’s user moves the cursor downward on the screen, the program steps to the next link to manipulate or display the new line. But what happens if the user moves the cursor upward? In an ordinary linked list, you would need to return current (or its equivalent) to the start of the list and then step all the way down again to the new current link. This isn’t very efficient. You want to make a single step upward.

The doubly linked list provides this capability. It allows you to traverse backward as well as forward through the list. The secret is that each link has two references to other links instead of one. The first is to the next link, as in ordinary lists. The second is to the previous link. This type of list is shown in Figure 5.13.

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next

next

next

next

first

Null

prev

prev

prev

prev

last

Null

**FIGURE 5.13 **

The beginning of the specification for the Link class in a doubly linked list looks like this:

class Link

{

public long dData; // data item

public Link next; // next link in list public link previous; // previous link in list

...

}

The downside of doubly linked lists is that every time you insert or delete a link you must deal with four links instead of two: two attachments to the previous link and two attachments to the following one. Also, of course, each link is a little bigger because of the extra reference.

A doubly linked list doesn’t necessarily need to be a double-ended list (keeping a reference to the last element on the list) but creating it this way is useful, so we’ll include it in our example.

We’ll show the complete listing for the doublyLinked.java program soon, but first let’s examine some of the methods in its doublyLinkedList class.

**Traversal**

Two display methods demonstrate traversal of a doubly linked list. The displayForward() method is the same as the displayList() method we’ve seen in ordinary linked lists. The displayBackward() method is similar but starts at the last element in the list and proceeds toward the start of the list, going to each element’s previous field. This code fragment shows how this process works: Link current = last; // start at end

while(current != null) // until start of list, current = current.previous; // move to previous link

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Incidentally, some people take the view that, because you can go either way equally easily on a doubly linked list, there is no preferred direction and therefore terms like previous and next are inappropriate. If you prefer, you can substitute direction-neutral terms such as left and right.

**Insertion**

We’ve included several insertion routines in the DoublyLinkedList class. The insertFirst() method inserts at the beginning of the list, insertLast() inserts at the end, and insertAfter() inserts following an element with a specified key.

Unless the list is empty, the insertFirst() routine changes the previous field in the old first link to point to the new link and changes the next field in the new link to point to the old first link. Finally, it sets first to point to the new link. This process is shown in Figure 5.14.

Old first link

next

next

next

next

first

Null

prev

prev

prev

prev

last

❸

❶

❷

next

prev

Null

New Link

**FIGURE 5.14 **

If the list is empty, the last field must be changed instead of the first.previous field.

Here’s the code:

if( isEmpty() ) // if empty list,

last = newLink; // newLink <-- last

else

first.previous = newLink; // newLink <-- old first newLink.next = first; // newLink --> old first first = newLink; // first --> newLink

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The insertLast() method is the same process applied to the end of the list; it’s a mirror image of insertFirst().

The insertAfter() method inserts a new link following the link with a specified key value. It’s a bit more complicated because four connections must be made. First, the link with the specified key value must be found. This procedure is handled the same way as the find() routine in the linkList2.java program (Listing 5.2). Then, assuming we’re not at the end of the list, two connections must be made between the new link and the next link, and two more between current and the new link. This process is shown in Figure 5.15.

current

next

next

next

next

first

Null

prev

prev

prev

prev

last

Null

❷

❹

❶

next

prev

❸

**FIGURE 5.15 **

If the new link will be inserted at the end of the list, its next field must point to null, and last must point to the new link. Here’s the insertAfter() code that deals with the links:

if(current==last) // if last link,

{

newLink.next = null; // newLink --> null

last = newLink; // newLink <-- last

}

else // not last link,

{

newLink.next = current.next; // newLink --> old next

// newLink <-- old next

current.next.previous = newLink;

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}

newLink.previous = current; // old current <-- newLink current.next = newLink; // old current --> newLink Perhaps you’re unfamiliar with the use of two dot operators in the same expression.

It’s a natural extension of a single dot operator. The expression current.next.previous

means the previous field of the link referred to by the next field in the link current.

**Deletion**

There are three deletion routines: deleteFirst(), deleteLast(), and deleteKey(). The first two are fairly straightforward. In deleteKey(), the key being deleted is current.

Assuming the link to be deleted is neither the first nor the last one in the list, the next field of current.previous (the link before the one being deleted) is set to point to current.next (the link following the one being deleted), and the previous field of current.next is set to point to current.previous. This disconnects the current link from the list. Figure 5.16 shows how this disconnection looks, and the following two statements carry it out:

current.previous.next = current.next;

current.next.previous = current.previous;

current.prev

current

current.next

next

next

next

first

❶

Null

prev

prev

prev

last

Null

❷

**FIGURE 5.16 **

Special cases arise if the link to be deleted is either the first or last in the list because first or last must be set to point to the next or the previous link. Here’s the code from deleteKey() for dealing with link connections: if(current==first) // first item?

first = current.next; // first --> old next else // not first

// old previous --> old next

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current.previous.next = current.next;

if(current==last) // last item?

last = current.previous; // old previous <-- last else // not last

// old previous <-- old next

current.next.previous = current.previous;

**The **doublyLinked.java **Program**

Listing 5.8 shows the complete doublyLinked.java program, which includes all the routines just discussed.

**LISTING 5.8**

The doublyLinked.java Program

// doublyLinked.java

// demonstrates doubly-linked list

// to run this program: C>java DoublyLinkedApp

////////////////////////////////////////////////////////////////

class Link

{

public long dData; // data item

public Link next; // next link in list public Link previous; // previous link in list

// -------------------------------------------------------------

public Link(long d) // constructor

{ dData = d; }

// -------------------------------------------------------------

public void displayLink() // display this link

{ System.out.print(dData + “ “); }

// -------------------------------------------------------------

} // end class Link

////////////////////////////////////////////////////////////////

class DoublyLinkedList

{

private Link first; // ref to first item private Link last; // ref to last item

// -------------------------------------------------------------

public DoublyLinkedList() // constructor

{

first = null; // no items on list yet last = null;

}

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**LISTING 5.8**

Continued

// -------------------------------------------------------------

public boolean isEmpty() // true if no links

{ return first==null; }

// -------------------------------------------------------------

public void insertFirst(long dd) // insert at front of list

{

Link newLink = new Link(dd); // make new link

if( isEmpty() ) // if empty list,

last = newLink; // newLink <-- last

else

first.previous = newLink; // newLink <-- old first newLink.next = first; // newLink --> old first first = newLink; // first --> newLink

}

// -------------------------------------------------------------

public void insertLast(long dd) // insert at end of list

{

Link newLink = new Link(dd); // make new link

if( isEmpty() ) // if empty list,

first = newLink; // first --> newLink else

{

last.next = newLink; // old last --> newLink newLink.previous = last; // old last <-- newLink

}

last = newLink; // newLink <-- last

}

// -------------------------------------------------------------

public Link deleteFirst() // delete first link

{ // (assumes non-empty list) Link temp = first;

if(first.next == null) // if only one item

last = null; // null <-- last

else

first.next.previous = null; // null <-- old next first = first.next; // first --> old next return temp;

}

// -------------------------------------------------------------

public Link deleteLast() // delete last link

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**LISTING 5.8**

Continued

{ // (assumes non-empty list) Link temp = last;

if(first.next == null) // if only one item

first = null; // first --> null

else

last.previous.next = null; // old previous --> null last = last.previous; // old previous <-- last return temp;

}

// -------------------------------------------------------------

// insert dd just after key

public boolean insertAfter(long key, long dd)

{ // (assumes non-empty list) Link current = first; // start at beginning while(current.dData != key) // until match is found,

{

current = current.next; // move to next link

if(current == null)

return false; // didn’t find it

}

Link newLink = new Link(dd); // make new link

if(current==last) // if last link,

{

newLink.next = null; // newLink --> null

last = newLink; // newLink <-- last

}

else // not last link,

{

newLink.next = current.next; // newLink --> old next

// newLink <-- old next

current.next.previous = newLink;

}

newLink.previous = current; // old current <-- newLink current.next = newLink; // old current --> newLink return true; // found it, did insertion

}

// -------------------------------------------------------------

public Link deleteKey(long key) // delete item w/ given key

{ // (assumes non-empty list) Link current = first; // start at beginning

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**LISTING 5.8**

Continued

while(current.dData != key) // until match is found,

{

current = current.next; // move to next link

if(current == null)

return null; // didn’t find it

}

if(current==first) // found it; first item?

first = current.next; // first --> old next else // not first

// old previous --> old next

current.previous.next = current.next;

if(current==last) // last item?

last = current.previous; // old previous <-- last else // not last

// old previous <-- old next

current.next.previous = current.previous;

return current; // return value

}

// -------------------------------------------------------------

public void displayForward()

{

System.out.print(“List (first-->last): “);

Link current = first; // start at beginning while(current != null) // until end of list,

{

current.displayLink(); // display data

current = current.next; // move to next link

}

System.out.println(“”);

}

// -------------------------------------------------------------

public void displayBackward()

{

System.out.print(“List (last-->first): “);

Link current = last; // start at end

while(current != null) // until start of list,

{

current.displayLink(); // display data

current = current.previous; // move to previous link

}

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**LISTING 5.8**

Continued

System.out.println(“”);

}

// -------------------------------------------------------------

} // end class DoublyLinkedList

////////////////////////////////////////////////////////////////

class DoublyLinkedApp

{

public static void main(String[] args)

{ // make a new list

DoublyLinkedList theList = new DoublyLinkedList();

theList.insertFirst(22); // insert at front

theList.insertFirst(44);

theList.insertFirst(66);

theList.insertLast(11); // insert at rear

theList.insertLast(33);

theList.insertLast(55);

theList.displayForward(); // display list forward theList.displayBackward(); // display list backward theList.deleteFirst(); // delete first item

theList.deleteLast(); // delete last item

theList.deleteKey(11); // delete item with key 11

theList.displayForward(); // display list forward theList.insertAfter(22, 77); // insert 77 after 22

theList.insertAfter(33, 88); // insert 88 after 33

theList.displayForward(); // display list forward

} // end main()

} // end class DoublyLinkedApp

////////////////////////////////////////////////////////////////

In main() we insert some items at the beginning of the list and at the end, display the items going both forward and backward, delete the first and last items and the item with key 11, display the list again (forward only), insert two items using the insertAfter() method, and display the list again. Here’s the output:

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List (first-->last): 66 44 22 11 33 55

List (last-->first): 55 33 11 22 44 66

List (first-->last): 44 22 33

List (first-->last): 44 22 77 33 88

The deletion methods and the insertAfter() method assume that the list isn’t empty.

Although for simplicity we don’t show it in main(), isEmpty() should be used to verify that there’s something in the list before attempting such insertions and deletions.

**Doubly Linked List as Basis for Deques**

A doubly linked list can be used as the basis for a deque, mentioned in the preceding chapter. In a deque you can insert and delete at either end, and the doubly linked list provides this capability.

**Iterators**

We’ve seen how the user of a list can find a link with a given key using a find() method. The method starts at the beginning of the list and examines each link until it finds one matching the search key. Other operations we’ve looked at, such as deleting a specified link or inserting before or after a specified link, also involve searching through the list to find the specified link. However, these methods don’t give the user any control over the traversal to the specified item.

Suppose you wanted to traverse a list, performing some operation on certain links.

For example, imagine a personnel file stored as a linked list. You might want to increase the wages of all employees who were being paid minimum wage, without affecting employees already above the minimum. Or suppose that in a list of mail-order customers, you decided to delete all customers who had not ordered anything in six months.

In an array, such operations are easy because you can use an array index to keep track of your position. You can operate on one item, then increment the index to point to the next item, and see if that item is a suitable candidate for the operation.

However, in a linked list, the links don’t have fixed index numbers. How can we provide a list’s user with something analogous to an array index? You could repeatedly use find() to look for appropriate items in a list, but that approach requires many comparisons to find each link. It’s far more efficient to step from link to link, checking whether each one meets certain criteria and performing the appropriate operation if it does.

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**A Reference in the List Itself? **

As users of a list class, what we need is access to a reference that can point to any arbitrary link. This way, we can examine or modify the link. We should be able to increment the reference so we can traverse along the list, looking at each link in turn, and we should be able to access the link pointed to by the reference.

Assuming we create such a reference, where will it be installed? One possibility is to use a field in the list itself, called current or something similar. You could access a link using current and increment current to move to the next link.

One problem with this approach is that you might need more than one such reference, just as you often use several array indices at the same time. How many would be appropriate? There’s no way to know how many the user might need. Thus, it seems easier to allow the user to create as many such references as necessary. To make this possible in an object-oriented language, it’s natural to embed each reference in a class object. This object can’t be the same as the list class because there’s only one list object, so it is normally implemented as a separate class.

**An Iterator Class**

Objects containing references to items in data structures, used to traverse these structures, are commonly called

{

private Link current;

...

}

The current field contains a reference to the link the iterator currently points to.

(The term

To use such an iterator, the user might create a list and then create an iterator object associated with the list. Actually, as it turns out, letting the list create the iterator is easier, so it can pass the iterator certain information, such as a reference to its first field. Thus, we add a getIterator() method to the list class; this method returns a suitable iterator object to the user. Here’s some abbreviated code in main() that shows how the class user would invoke an iterator:

public static void main(...)

{

LinkList theList = new LinkList(); // make list ListIterator iter1 = theList.getIterator(); // make iter

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Link aLink = iter1.getCurrent(); // access link at iterator iter1.nextLink(); // move iter to next link

}

After we’ve made the iterator object, we can use it to access the link it points to or increment it so it points to the next link, as shown in the second two statements.

We call the iterator object iter1 to emphasize that you could make more iterators (iter2 and so on) the same way.

The iterator always points to some link in the list. It’s associated with the list, but it’s not the same as the list or the same as a link. Figure 5.17 shows two iterators pointing to links in a list.

Linked List

first

Null

next

next

next

next

List iterator 1

current

List iterator 2

current

**FIGURE 5.17 **

**Additional Iterator Features**

We’ve seen several programs in which the use of a previous field made performing certain operations simpler, such as deleting a link from an arbitrary location. Such a field is also useful in an iterator.

Also, it may be that the iterator will need to change the value of the list’s first field—for instance, if an item is inserted or deleted at the beginning of the list. If the iterator is an object of a separate class, how can it access a private field, such as first, in the list? One solution is for the list to pass a reference from itself to the iterator when it creates the iterator. This reference is stored in a field in the iterator.

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The list must then provide public methods that allow the iterator to change first.

These LinkList methods are getFirst() and setFirst(). (The weakness of this approach is that these methods allow anyone to change first, which introduces an element of risk.)

Here’s a revised (although still incomplete) iterator class that incorporates these additional fields, along with reset() and nextLink() methods: class ListIterator()

{

private Link current; // reference to current link private Link previous; // reference to previous link private LinkList ourList; // reference to “parent” list public void reset() // set to start of list

{

current = ourList.getFirst(); // current --> first previous = null; // previous --> null

}

public void nextLink() // go to next link

{

previous = current; // set previous to this

current = current.next; // set this to next

}

...

}

We might note, for you old-time C++ programmers, that in C++ the connection between the iterator and the list is typically provided by making the iterator class a

**Iterator Methods**

Additional methods can make the iterator a flexible and powerful class. All operations previously performed by the class that involve iterating through the list, such as insertAfter(), are more naturally performed by the iterator. In our example the iterator includes the following methods:

• reset()—Sets the iterator to the start of the list

• nextLink()—Moves the iterator to the next link

• getCurrent()—Returns the link at the iterator

• atEnd()—Returns true if the iterator is at the end of the list

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• insertAfter()—Inserts a new link after the iterator

• insertBefore()—Inserts a new link before the iterator

• deleteCurrent()—Deletes the link at the iterator

The user can position the iterator using reset() and nextLink(), check whether it’s at the end of the list with atEnd(), and perform the other operations shown.

Deciding which tasks should be carried out by an iterator and which by the list itself is not always easy. An insertBefore() method works best in the iterator, but an insertFirst() routine that always inserts at the beginning of the list might be more appropriate in the list class. We’ve kept a displayList() routine in the list, but this operation could also be handled with getCurrent() and nextLink() calls to the iterator.

**The **interIterator.java **Program**

The interIterator.java program includes an interactive interface that permits the user to control the iterator directly. After you’ve started the program, you can perform the following actions by typing the appropriate letter:

• s—Show the list contents

• r—Reset the iterator to the start of the list

• n—Go to the next link

• g—Get the contents of the current link

• b—Insert before the current link

• a—Insert a new link after the current link

• d—Delete the current link

Listing 5.9 shows the complete interIterator.java program.

**LISTING 5.9**

The interIterator.java Program

// interIterator.java

// demonstrates iterators on a linked listListIterator

// to run this program: C>java InterIterApp

import java.io.*; // for I/O

////////////////////////////////////////////////////////////////

class Link

{

public long dData; // data item

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**LISTING 5.9**

Continued

public Link next; // next link in list

// -------------------------------------------------------------

public Link(long dd) // constructor

{ dData = dd; }

// -------------------------------------------------------------

public void displayLink() // display ourself

{ System.out.print(dData + “ “); }

} // end class Link

////////////////////////////////////////////////////////////////

class LinkList

{

private Link first; // ref to first item on list

// -------------------------------------------------------------

public LinkList() // constructor

{ first = null; } // no items on list yet

// -------------------------------------------------------------

public Link getFirst() // get value of first

{ return first; }

// -------------------------------------------------------------

public void setFirst(Link f) // set first to new link

{ first = f; }

// -------------------------------------------------------------

public boolean isEmpty() // true if list is empty

{ return first==null; }

// -------------------------------------------------------------

public ListIterator getIterator() // return iterator

{

return new ListIterator(this); // initialized with

} // this list

// -------------------------------------------------------------

public void displayList()

{

Link current = first; // start at beginning of list while(current != null) // until end of list,

{

current.displayLink(); // print data

current = current.next; // move to next link

}

System.out.println(“”);

}

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**LISTING 5.9**

Continued

// -------------------------------------------------------------

} // end class LinkList

////////////////////////////////////////////////////////////////

class ListIterator

{

private Link current; // current link

private Link previous; // previous link

private LinkList ourList; // our linked list

//--------------------------------------------------------------

public ListIterator(LinkList list) // constructor

{

ourList = list;

reset();

}

//--------------------------------------------------------------

public void reset() // start at ‘first’

{

current = ourList.getFirst();

previous = null;

}

//--------------------------------------------------------------

public boolean atEnd() // true if last link

{ return (current.next==null); }

//--------------------------------------------------------------

public void nextLink() // go to next link

{

previous = current;

current = current.next;

}

//--------------------------------------------------------------

public Link getCurrent() // get current link

{ return current; }

//--------------------------------------------------------------

public void insertAfter(long dd) // insert after

{ // current link

Link newLink = new Link(dd);

if( ourList.isEmpty() ) // empty list

{

ourList.setFirst(newLink);

current = newLink;

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**CHAPTER 5**

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**LISTING 5.9**

Continued

}

else // not empty

{

newLink.next = current.next;

current.next = newLink;

nextLink(); // point to new link

}

}

//--------------------------------------------------------------

public void insertBefore(long dd) // insert before

{ // current link

Link newLink = new Link(dd);

if(previous == null) // beginning of list

{ // (or empty list)

newLink.next = ourList.getFirst();

ourList.setFirst(newLink);

reset();

}

else // not beginning

{

newLink.next = previous.next;

previous.next = newLink;

current = newLink;

}

}

//--------------------------------------------------------------

public long deleteCurrent() // delete item at current

{

long value = current.dData;

if(previous == null) // beginning of list

{

ourList.setFirst(current.next);

reset();

}

else // not beginning

{

previous.next = current.next;

if( atEnd() )

reset();

else

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**LISTING 5.9**

Continued

current = current.next;

}

return value;

}

//--------------------------------------------------------------

} // end class ListIterator

////////////////////////////////////////////////////////////////

class InterIterApp

{

public static void main(String[] args) throws IOException

{

LinkList theList = new LinkList(); // new list ListIterator iter1 = theList.getIterator(); // new iter long value;

iter1.insertAfter(20); // insert items

iter1.insertAfter(40);

iter1.insertAfter(80);

iter1.insertBefore(60);

while(true)

{

System.out.print(“Enter first letter of show, reset, “); System.out.print(“next, get, before, after, delete: “); System.out.flush();

int choice = getChar(); // get user’s option switch(choice)

{

case ‘s’: // show list

if( !theList.isEmpty() )

theList.displayList();

else

System.out.println(“List is empty”);

break;

case ‘r’: // reset (to first)

iter1.reset();

break;

case ‘n’: // advance to next item if( !theList.isEmpty() && !iter1.atEnd() )

iter1.nextLink();

else

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**LISTING 5.9**

Continued

System.out.println(“Can’t go to next link”);

break;

case ‘g’: // get current item

if( !theList.isEmpty() )

{

value = iter1.getCurrent().dData;

System.out.println(“Returned “ + value);

}

else

System.out.println(“List is empty”);

break;

case ‘b’: // insert before current System.out.print(“Enter value to insert: “);

System.out.flush();

value = getInt();

iter1.insertBefore(value);

break;

case ‘a’: // insert after current System.out.print(“Enter value to insert: “);

System.out.flush();

value = getInt();

iter1.insertAfter(value);

break;

case ‘d’: // delete current item if( !theList.isEmpty() )

{

value = iter1.deleteCurrent();

System.out.println(“Deleted “ + value);

}

else

System.out.println(“Can’t delete”);

break;

default:

System.out.println(“Invalid entry”);

} // end switch

} // end while

} // end main()

//--------------------------------------------------------------

public static String getString() throws IOException

{

InputStreamReader isr = new InputStreamReader(System.in);

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**LISTING 5.9**

Continued

BufferedReader br = new BufferedReader(isr);

String s = br.readLine();

return s;

}

//-------------------------------------------------------------

public static char getChar() throws IOException

{

String s = getString();

return s.charAt(0);

}

//-------------------------------------------------------------

public static int getInt() throws IOException

{

String s = getString();

return Integer.parseInt(s);

}

//-------------------------------------------------------------

} // end class InterIterApp

////////////////////////////////////////////////////////////////

The main() routine inserts four items into the list, using an iterator and its insertAfter() method. Then it waits for the user to interact with it. In the following sample interaction, the user displays the list, resets the iterator to the beginning, goes forward two links, gets the current link’s key value (which is 60), inserts 100

before this, inserts 7 after the 100, and displays the list again: Enter first letter of

show, reset, next, get, before, after, delete: s

20 40 60 80

Enter first letter of

show, reset, next, get, before, after, delete: r

Enter first letter of

show, reset, next, get, before, after, delete: n

Enter first letter of

show, reset, next, get, before, after, delete: n

Enter first letter of

show, reset, next, get, before, after, delete: g

Returned 60

Enter first letter of

show, reset, next, get, before, after, delete: b

Enter value to insert: 100

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Enter first letter of

show, reset, next, get, before, after, delete: a

Enter value to insert: 7

Enter first letter of

show, reset, next, get, before, after, delete: s

20 40 100 7 60 80

Experimenting with the interIterator.java program will give you a feeling for how the iterator moves along the links and how it can insert and delete links anywhere in the list.

**Where Does the Iterator Point? **

One of the design issues in an iterator class is deciding where the iterator should point following various operations.

When you delete an item with deleteCurrent(), should the iterator end up pointing to the next item, to the previous item, or back at the beginning of the list? Keeping the iterator in the vicinity of the deleted item is convenient because the chances are the class user will be carrying out other operations there. However, you can’t move it to the previous item because there’s no way to reset the list’s previous field to the previous item. (You would need a doubly linked list for that task.) Our solution is to move the iterator to the link following the deleted link. If we’ve just deleted the item at the end of the list, the iterator is set to the beginning of the list.

Following calls to insertBefore() and insertAfter(), we return with current pointing to the newly inserted item.

**The **atEnd() **Method**

There’s another question about the atEnd() method. It could return true when the iterator points to the last valid link in the list, or it could return true when the iterator points

With the first approach, a loop condition used to iterate through the list becomes awkward because you need to perform an operation on the last link before checking whether it is the last link (and terminating the loop if it is).

However, the second approach doesn’t allow you to find out you’re at the end of the list until it’s too late to do anything with the last link. (You couldn’t look for the last link and then delete it, for example.) This is because when atEnd() became true, the iterator would no longer point to the last link (or indeed any valid link), and you can’t “back up” the iterator in a singly linked list.

We take the first approach. This way, the iterator always points to a valid link, although you must be careful when writing a loop that iterates through the list, as we’ll see next.

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**Iterative Operations**

As we noted, an iterator allows you to traverse the list, performing operations on certain data items. Here’s a code fragment that displays the list contents, using an iterator instead of the list’s displayList() method: iter1.reset(); // start at first long value = iter1.getCurrent().dData; // display link System.out.println(value + “ “);

while( !iter1.atEnd() ) // until end,

{

iter1.nextLink(); // go to next link, long value = iter1.getCurrent().dData; // display it System.out.println(value + “ “);

}

Although we don’t do so here, you should check with isEmpty() to be sure the list is not empty before calling getCurrent().

The following code shows how you could delete all items with keys that are multiples of 3. We show only the revised main() routine; everything else is the same as in interIterator.java (Listing 5.9).

class InterIterApp

{

public static void main(String[] args) throws IOException

{

LinkList theList = new LinkList(); // new list ListIterator iter1 = theList.getIterator(); // new iter iter1.insertAfter(21); // insert links

iter1.insertAfter(40);

iter1.insertAfter(30);

iter1.insertAfter(7);

iter1.insertAfter(45);

theList.displayList(); // display list

iter1.reset(); // start at first link Link aLink = iter1.getCurrent(); // get it

if(aLink.dData % 3 == 0) // if divisible by 3, iter1.deleteCurrent(); // delete it

while( !iter1.atEnd() ) // until end of list,

{

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iter1.nextLink(); // go to next link

aLink = iter1.getCurrent(); // get link

if(aLink.dData % 3 == 0) // if divisible by 3, iter1.deleteCurrent(); // delete it

}

theList.displayList(); // display list

} // end main()

} // end class InterIterApp

We insert five links and display the list. Then we iterate through the list, deleting those links with keys divisible by 3, and display the list again. Here’s the output: 21 40 30 7 45

40 7

Again, although we don’t show it here, it’s important to check whether the list is empty before calling deleteCurrent().

**Other Methods**

You could create other useful methods for the ListIterator class. For example, a find() method would return an item with a specified key value, as we’ve seen when find() is a list method. A replace() method could replace items that had certain key values with other items.

Because it’s a singly linked list, you can iterate along it only in the forward direction.

If a doubly linked list were used, you could go either way, allowing operations such as deletion from the end of the list, just as with non-iterators. This capability would probably be a convenience in some applications.

**Summary**

• A linked list consists of one linkedList object and a number of Link objects.

• The linkedList object contains a reference, often called first, to the first link in the list.

• Each Link object contains data and a reference, often called next, to the next link in the list.

• A next value of null signals the end of the list.

• Inserting an item at the beginning of a linked list involves changing the new link’s next field to point to the old first link and changing first to point to the new item.

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• Deleting an item at the beginning of a list involves setting first to point to first.next.

• To traverse a linked list, you start at first and then go from link to link, using each link’s next field to find the next link.

• A link with a specified key value can be found by traversing the list. Once found, an item can be displayed, deleted, or operated on in other ways.

• A new link can be inserted before or after a link with a specified key value, following a traversal to find this link.

• A double-ended list maintains a pointer to the last link in the list, often called last, as well as to the first.

• A double-ended list allows insertion at the end of the list.

• An Abstract Data Type (ADT) is a data storage class considered without reference to its implementation.

• Stacks and queues are ADTs. They can be implemented using either arrays or linked lists.

• In a sorted linked list, the links are arranged in order of ascending (or sometimes descending) key value.

• Insertion in a sorted list takes O(N) time because the correct insertion point must be found. Deletion of the smallest link takes O(1) time.

• In a doubly linked list, each link contains a reference to the previous link as well as the next link.

• A doubly linked list permits backward traversal and deletion from the end of the list.

• An iterator is a reference, encapsulated in a class object, that points to a link in an associated list.

• Iterator methods allow the user to move the iterator along the list and access the link currently pointed to.

• An iterator can be used to traverse through a list, performing some operation on selected links (or all links).

**Questions**

These questions are intended as a self-test for readers. Answers may be found in Appendix C.

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**1. **Which of the following is **a. **can be used to access public methods in the object.

**b. **has a size dependant on its class.

**c. **has the data type of the class.

**d. **does not hold the object itself.

**2. **Access to the links in a linked list is usually through the _________ link.

**3. **When you create a reference to a link in a linked list, it **a. **must refer to the first link.

**b. **must refer to the link pointed to by current.

**c. **must refer to the link pointed to by next.

**d. **can refer to any link you want.

**4. **How many references must you change to insert a link in the middle of a singly linked list?

**5. **How many references must you change to insert a link at the end of a singly linked list?

**6. **In the insertFirst() method in the linkList.java program (Listing 5.1), the statement newLink.next=first; means that

**a. **the next new link to be inserted will refer to first.

**b. **first will refer to the new link.

**c. **the next field of the new link will refer to the old first link.

**d. **newLink.next will refer to the new first link in the list.

**7. **Assuming current points to the next-to-last link in a singly linked list, what statement will delete the last link from the list?

**8. **When all references to a link are changed to refer to something else, what happens to the link?

**9. **A double-ended list

**a. **can be accessed from either end.

**b. **is a different name for a doubly linked list.

**c. **has pointers running both forward and backward between links.

**d. **has its first link connected to its last link.

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**10. **A special case often occurs for insertion and deletion routines when a list is ________.

**11. **Assuming a copy takes longer than a comparison, is it faster to delete an item with a certain key from a linked list or from an unsorted array?

**12. **How many times would you need to traverse a singly linked list to delete the item with the largest key?

**13. **Of the lists discussed in this chapter, which one would be best for implementing a queue?

**14. **Which of the following is **a. **do an insertion sort on a linked list.

**b. **insert a new link at the beginning of a list.

**c. **swap two links at arbitrary locations.

**d. **delete all links with a certain key value.

**15. **Which do you think would be a better choice to implement a stack: a singly linked list or an array?

**Experiments**

**1. **Use the LinkList Workshop applet to execute insert, find, and delete operations on both sorted and unsorted lists. For the operations demonstrated by this applet, is there any advantage to the sorted list?

**2. **Modify main() in the linkList.java program (Listing 5.1) so that it continuously inserts links into the list until memory is exhausted. After each 1,000

items, have it display the number of items inserted so far. This way, you can learn approximately how many links a list can hold in your particular machine. (Of course, the number will vary depending on what other programs are in memory and many other factors.) Don’t try this experiment if it will crash your institution’s network.

**Programming Projects**

Writing programs that solve the Programming Projects helps to solidify your understanding of the material and demonstrates how the chapter’s concepts are applied.

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(As noted in the Introduction, qualified instructors may obtain completed solutions to the Programming Projects on the publisher’s Web site.) **5.1 **Implement a priority queue based on a sorted linked list. The remove operation on the priority queue should remove the item with the smallest key.

**5.2 **Implement a deque based on a doubly linked list. (See Programming Project 4.2

in the preceding chapter.) The user should be able to carry out the normal operations on the deque.

**5.3 **A circular list is a linked list in which the last link points back to the first link.

There are many ways to design a circular list. Sometimes there is a pointer to the “start” of the list. However, this makes the list less like a real circle and more like an ordinary list that has its end attached to its beginning. Make a class for a singly linked circular list that has no end and no beginning. The only access to the list is a single reference, current, that can point to any link on the list. This reference can move around the list as needed. (See Programming Project 5.5 for a situation in which such a circular list is ideally suited.) Your list should handle insertion, searching, and deletion. You may find it convenient if these operations take place one link downstream of the link pointed to by current. (Because the upstream link is singly linked, you can’t get at it without going all the way around the circle.) You should also be able to display the list (although you’ll need to break the circle at some arbitrary point to print it on the screen). A step() method that moves current along to the next link might come in handy too.

**5.4 **Implement a stack class based on the circular list of Programming Project 5.3.

This exercise is not too difficult. (However, implementing a queue can be harder, unless you make the circular list doubly linked.) **5.5 **The Josephus Problem is a famous mathematical puzzle that goes back to ancient times. There are many stories to go with the puzzle. One is that Josephus was one of a group of Jews who were about to be captured by the Romans. Rather than be enslaved, they chose to commit suicide. They arranged themselves in a circle and, starting at a certain person, started counting off around the circle. Every nth person had to leave the circle and commit suicide.

Josephus decided he didn’t want to die, so he arranged the rules so he would be the last person left. If there were (say) 20 people, and he was the seventh person from the start of the circle, what number should he tell them to use for counting off? The problem is made much more complicated because the circle shrinks as the counting continues.

Create an application that uses a circular linked list (like that in Programming Project 5.3) to model this problem. Inputs are the number of people in the circle, the number used for counting off, and the number of the person where

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249

counting starts (usually 1). The output is the list of persons being eliminated.

When a person drops out of the circle, counting starts again from the person who was on his left (assuming you go around clockwise). Here’s an example.

There are seven people numbered 1 through 7, and you start at 1 and count off by threes. People will be eliminated in the order 4, 1, 6, 5, 7, 3. Number 2 will be left.

**5.6 **Let’s try something a little different: a two-dimensional linked list, which we’ll call a matrix. This is the list analogue of a two-dimensional array. It might be useful in applications such as spreadsheet programs. If a spreadsheet is based on an array, and you insert a new row near the top, you must move every cell in the lower rows N*M cells, which is potentially a slow process. If the spreadsheet is implemented by a matrix, you need only change N pointers.

For simplicity, we’ll assume a singly linked approach (although a double-linked approach would probably be more appropriate for a spreadsheet). Each link (except those on the top row and left side) is pointed to by the link directly above it and by the link on its left. You can start at the upper-left link and navigate to, say, the link on the third row and fifth column by following the pointers down two rows and right four columns. Assume your matrix is created with specified dimensions (7 by 10, for example). You should be able to insert values in specified links and display the contents of the matrix.

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**6**

**IN THIS CHAPTER**

• Triangular Numbers

Recursion

• Factorials

• Anagrams

• A Recursive Binary Search

Recursion is a programming technique in which a

method (function) calls itself. This may sound like a

• The Towers of Hanoi

strange thing to do, or even a catastrophic mistake.

•

Recursion is, however, one of the most interesting, and Mergesort

one of the most surprisingly effective, techniques in

• Eliminating Recursion

programming. Like pulling yourself up by your bootstraps (you do have bootstraps, don’t you?), recursion seems

• Some Interesting Recursive

incredible when you first encounter it. However, it not Applications

only works, it also provides a unique conceptual frame-work for solving many problems.

In this chapter we’ll examine numerous examples to show the wide variety of situations to which recursion can be applied. We will calculate triangular numbers and factorials, generate anagrams, perform a recursive binary search, solve the Towers of Hanoi puzzle, and investigate a sorting technique called mergesort. Workshop applets are provided to demonstrate the Towers of Hanoi and mergesort.

We’ll also discuss the strengths and weaknesses of recursion, and show how a recursive approach can be transformed into a stack-based approach.

**Triangular Numbers**

It’s said that the Pythagorians, a band of mathematicians in ancient Greece who worked under Pythagoras (of

Pythagorian theorem fame), felt a mystical connection with the series of numbers 1, 3, 6, 10, 15, 21, … (where the

… means the series continues indefinitely). Can you find the next member of this series?

The nth term in the series is obtained by adding n to the previous term. Thus, the second term is found by adding 2

to the first term (which is 1), giving 3. The third term is 3

added to the second term (which is 3) giving 6, and so on.

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**CHAPTER 6**

Recursion

The numbers in this series are called

#1 = 1

#2 = 3

#3 = 6

#4 = 10

#5 = 15

#6 = 21

#7 = 28

**FIGURE 6.1 **

**Finding the **n**th Term Using a Loop**

Suppose you wanted to find the value of some arbitrary nth term in the series—say the fourth term (whose value is 10). How would you calculate it? Looking at Figure 6.2, you might decide that the value of any term can be obtained by adding up all the vertical columns of squares.

1 in this column

2 in this column

3 in this column

4 in this column

Total: 10

**FIGURE 6.2 **

In the fourth term, the first column has four little squares, the second column has three, and so on. Adding 4+3+2+1 gives 10.

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The following triangle() method uses this column-based technique to find a triangular number. It sums all the columns, from a height of n to a height of 1: int triangle(int n)

{

int total = 0;

while(n > 0) // until n is 1

{

total = total + n; // add n (column height) to total

--n; // decrement column height

}

return total;

}

The method cycles around the loop n times, adding n to total the first time, n-1 the second time, and so on down to 1, quitting the loop when n becomes 0.

**Finding the **n**th Term Using Recursion**

The loop approach may seem straightforward, but there’s another way to look at this problem. The value of the nth term can be thought of as the sum of only two things, instead of a whole series. They are

**1. **The first (tallest) column, which has the value n.

**2. **The sum of all the remaining columns.

This is shown in Figure 6.3.

6 in the remaining columns

4 in the first column

Total: 10

**FIGURE 6.3 **

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**Finding the Remaining Columns**

If we knew about a method that found the sum of all the remaining columns, we could write our triangle() method, which returns the value of the nth triangular number, like this:

int triangle(int n)

{

return( n + sumRemainingColumns(n) ); // (incomplete version)

}

But what have we gained here? It looks like writing the sumRemainingColumns() method is just as hard as writing the triangle() method in the first place.

Notice in Figure 6.3, however, that the sum of all the remaining columns for term n is the same as the sum of

{

return( n + sumAllColumns(n-1) ); // (incomplete version)

}

But when you think about it, the sumAllColumns() method is doing exactly the same thing the triangle() method is: summing all the columns for some number n passed as an argument. So why not use the triangle() method itself, instead of some other method? That would look like this:

int triangle(int n)

{

return( n + triangle(n-1) ); // (incomplete version)

}

You may be amazed that a method can call itself, but why shouldn’t it be able to? A method call is (among other things) a transfer of control to the start of the method.

This transfer of control can take place from within the method as well as from outside.

**Passing the Buck**

All these approaches may seem like passing the buck. Someone tells me to find the 9th triangular number. I know this is 9 plus the 8th triangular number, so I call Harry and ask him to find the 8th triangular number. When I hear back from him, I’ll add 9 to whatever he tells me, and that will be the answer.

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Harry knows the 8th triangular number is 8 plus the 7th triangular number, so he calls Sally and asks her to find the 7th triangular number. This process continues with each person passing the buck to another one.

Where does this buck-passing end? Someone at some point must be able to figure out an answer that doesn’t involve asking another person to help. If this didn’t happen, there would be an infinite chain of people asking other people questions—a sort of arithmetic Ponzi scheme that would never end. In the case of triangle(), this would mean the method calling itself over and over in an infinite series that would eventually crash the program.

**The Buck Stops Here**

To prevent an infinite regress, the person who is asked to find the first triangular number of the series, when n is 1, must know, without asking anyone else, that the answer is 1. There are no smaller numbers to ask anyone about, there’s nothing left to add to anything else, so the buck stops there. We can express this by adding a condition to the triangle() method:

int triangle(int n)

{

if(n==1)

return 1;

else

return( n + triangle(n-1) );

}

The condition that leads to a recursive method returning without making another recursive call is referred to as the

**The **triangle.java **Program**

Does recursion actually work? If you run the triangle.java program, you’ll see that it does. Enter a value for the term number, n, and the program will display the value of the corresponding triangular number. Listing 6.1 shows the triangle.java program.

**LISTING 6.1**

The triangle.java Program

// triangle.java

// evaluates triangular numbers

// to run this program: C>java TriangleApp

import java.io.*; // for I/O

////////////////////////////////////////////////////////////////

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**CHAPTER 6**

Recursion

**LISTING 6.1**

Continued

class TriangleApp

{

static int theNumber;

public static void main(String[] args) throws IOException

{

System.out.print(“Enter a number: “);

theNumber = getInt();

int theAnswer = triangle(theNumber);

System.out.println(“Triangle=”+theAnswer);

} // end main()

//-------------------------------------------------------------

public static int triangle(int n)

{

if(n==1)

return 1;

else

return( n + triangle(n-1) );

}

//-------------------------------------------------------------

public static String getString() throws IOException

{

String s = br.readLine();

return s;

}

//--------------------------------------------------------------

public static int getInt() throws IOException

{

String s = getString();

return Integer.parseInt(s);

}

//--------------------------------------------------------------

} // end class TriangleApp

////////////////////////////////////////////////////////////////

The main() routine prompts the user for a value for n, calls triangle(), and displays the return value. The triangle() method calls itself repeatedly to do all the work.

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257

Here’s some sample output:

Enter a number: 1000

Triangle = 500500

Incidentally, if you’re skeptical of the results returned from triangle(), you can check them by using the following formula:

nth triangular number = (n2+n)/2

**What’s Really Happening? **

Let’s modify the triangle() method to provide an insight into what’s happening when it executes. We’ll insert some output statements to keep track of the arguments and return values:

public static int triangle(int n)

{

System.out.println(“Entering: n=” + n);

if(n==1)

{

System.out.println(“Returning 1”);

return 1;

}

else

{

int temp = n + triangle(n-1);

System.out.println(“Returning “ + temp);

return temp;

}

}

Here’s the interaction when this method is substituted for the earlier triangle() method and the user enters 5:

Enter a number: 5

Entering: n=5

Entering: n=4

Entering: n=3

Entering: n=2

Entering: n=1

Returning 1

Returning 3

Returning 6

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Returning 10

Returning 15

Triangle = 15

Each time the triangle() method calls itself, its argument, which starts at 5, is reduced by 1. The method plunges down into itself again and again until its argument is reduced to 1. Then it returns. This triggers an entire series of returns. The method rises back up, phoenix-like, out of the discarded versions of itself. Each time it returns, it adds the value of n it was called with to the return value from the method it called.

The return values recapitulate the series of triangular numbers, until the answer is returned to main(). Figure 6.4 shows how each invocation of the triangle() method can be imagined as being “inside” the previous one.

called with n=5

Version 1

n=5

Version 2

n=4

Version 3

n=3

Version 4

n=2

Version 5

n=1

Returns 1

Adds 2

Returns 3

Adds 3

Returns 6

Adds 4

Returns 10

Adds 5

Returns 15

Returns 15

**FIGURE 6.4**

The recursive triangle() method.

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Notice that, just before the innermost version returns a 1, there are actually five different incarnations of triangle() in existence at the same time. The outer one was passed the argument 5; the inner one was passed the argument 1.

**Characteristics of Recursive Methods**

Although it’s short, the triangle() method possesses the key features common to all recursive routines:

• It calls itself.

• When it calls itself, it does so to solve a smaller problem.

• There’s some version of the problem that is simple enough that the routine can solve it, and return, without calling itself.

In each successive call of a recursive method to itself, the argument becomes smaller (or perhaps a range described by multiple arguments becomes smaller), reflecting the fact that the problem has become “smaller” or easier. When the argument or range reaches a certain minimum size, a condition is triggered and the method returns without calling itself.

**Is Recursion Efficient? **

Calling a method involves certain overhead. Control must be transferred from the location of the call to the beginning of the method. In addition, the arguments to the method and the address to which the method should return must be pushed onto an internal stack so that the method can access the argument values and know where to return.

In the case of the triangle() method, it’s probable that, as a result of this overhead, the while loop approach executes more quickly than the recursive approach. The penalty may not be significant, but if there are a large number of method calls as a result of a recursive method, it might be desirable to eliminate the recursion. We’ll talk about this issue more at the end of this chapter.

Another inefficiency is that memory is used to store all the intermediate arguments and return values on the system’s internal stack. This may cause problems if there is a large amount of data, leading to stack overflow.

Recursion is usually used because it simplifies a problem conceptually, not because it’s inherently more efficient.

**Mathematical Induction**

Recursion is the programming equivalent of mathematical induction. Mathematical induction is a way of defining something in terms of itself. (The term is also used to

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describe a related approach to proving theorems.) Using induction, we could define the triangular numbers mathematically by saying

tri(n) = 1

if n = 1

tri(n) = n + tri(n–1)

if n > 1

Defining something in terms of itself may seem circular, but in fact it’s perfectly valid (provided there’s a base case).

**Factorials**

Factorials are similar in concept to triangular numbers, except that multiplication is used instead of addition. The triangular number corresponding to n is found by adding n to the triangular number of n-1, while the factorial of n is found by multiplying n by the factorial of n-1. That is, the fifth triangular number is 5+4+3+2+1, while the factorial of 5 is 5*4*3*2*1, which equals 120. Table 6.1 shows the factorials of the first 10 numbers.

**TABLE 6.1**

Factorials

**Number**

**Calculation**

**Factorial**

0

by definition

1

1

1 * 1

1

2

2 * 1

2

3

3 * 2

6

4

4 * 6

24

5

5 * 24

120

6

6 * 120

720

7

7 * 720

5,040

8

8 * 5,040

40,320

9

9 * 40,320

362,880

The factorial of 0 is defined to be 1. Factorial numbers grow large very rapidly, as you can see.

A recursive method similar to triangle() can be used to calculate factorials. It looks like this:

int factorial(int n)

{

if(n==0)

return 1;

else

return (n * factorial(n-1) );

}

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There are only two differences between factorial() and triangle(). First, factorial() uses a * instead of a + in the expression

n * factorial(n-1)

Second, the base condition occurs when n is 0, not 1. Here’s some sample interaction when this method is used in a program similar to triangle.java: Enter a number: 6

Factorial =720

Figure 6.5 shows how the various incarnations of factorial() call themselves when initially entered with n=4.

called with n=4

Version 1

n=4

Version 2

n=3

Version 3

n=2

Version 4

n=1

Version 5

n=0

Return 1

Multiply by 1

Return 1

Multiply by 2

Return 2

Multiply by 3

Return 6

Multiply by 4

Return 24

Returns 24

**FIGURE 6.5**

The recursive factorial() method.

Calculating factorials is the classic demonstration of recursion, although factorials aren’t as easy to visualize as triangular numbers.

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Various other numerological entities lend themselves to calculation using recursion in a similar way, such as finding the greatest common denominator of two numbers (which is used to reduce a fraction to lowest terms), raising a number to a power, and so on. Again, while these calculations are interesting for demonstrating recursion, they probably wouldn’t be used in practice because a loop-based approach is more efficient.

**Anagrams**

Here’s a different kind of situation in which recursion provides a neat solution to a problem. A permutation is an arrangement of things in a definite order. Suppose you want to list all the anagrams of a specified word—that is, all possible permutations (whether they make a real English word or not) that can be made from the letters of the original word. We’ll call this

• cat

• cta

• atc

• act

• tca

• tac

Try anagramming some words yourself. You’ll find that the number of possibilities is the factorial of the number of letters. For 3 letters there are 6 possible words; for 4

letters there are 24 words; for 5 letters, 120; and so on. (This assumes that all letters are distinct; if there are multiple instances of the same letter, there will be fewer possible words.)

How would you write a program to anagram a word? Here’s one approach. Assume the word has n letters.

**1. **Anagram the rightmost n-1 letters.

**2. **Rotate all n letters.

**3. **Repeat these steps n times.

To

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Temp

Word

r

o

d

e

o

r

o

d

e

o

r

o

d

e

o

o

d

e

o

r

**FIGURE 6.6**

Rotating a word.

Rotating the word n times gives each letter a chance to begin the word. While the selected letter occupies this first position, all the other letters are then anagrammed (arranged in every possible position). For cat, which has only three letters, rotating the remaining two letters simply switches them. The sequence is shown in Table 6.2.

**TABLE 6.2**

Anagramming the Word cat

**Word**

**Display**

**First**

**Remaining**

**Action**

**Word? **

**Letter**

**Letters**

cat

Yes

c

at

Rotate at

cta

Yes

c

ta

Rotate ta

cat

No

c

at

Rotate cat

atc

Yes

a

tc

Rotate tc

act

Yes

a

ct

Rotate ct

atc

No

a

tc

Rotate atc

tca

Yes

t

ca

Rotate ca

tac

Yes

t

ac

Rotate ac

tca

No

t

ca

Rotate tca

cat

No

c

at

Done

Notice that we must rotate back to the starting point with two letters before performing a three-letter rotation. This leads to sequences like cat, cta, cat. The redundant sequences aren’t displayed.

How do we anagram the rightmost n-1 letters? By calling ourselves. The recursive doAnagram() method takes the size of the word to be anagrammed as its only parameter. This word is understood to be the rightmost n letters of the complete word. Each

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time doAnagram() calls itself, it does so with a word one letter smaller than before, as shown in Figure 6.7.

called with

word = cat

Word = cat

Word = at

Word = t

Display cat

Rotate at

Word = a

Display cta

Rotate ta

Rotate cat = atc

Word = tc

Word = c

Display atc

Rotate tc

Word = t

Display act

Rotate ct

Rotate atc = tca

Word = ca

Word = a

Rotate ca

Display tca

Word = c

Rotate ac

Display tac

Rotate tca = cat

**FIGURE 6.7**

The recursive doAnagram() method.

The base case occurs when the size of the word to be anagrammed is only one letter.

There’s no way to rearrange one letter, so the method returns immediately.

Otherwise, it anagrams all but the first letter of the word it was given and then rotates the entire word. These two actions are performed n times, where n is the size of the word. Here’s the recursive routine doAnagram():

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public static void doAnagram(int newSize)

{

if(newSize == 1) // if too small, return; // go no further

for(int j=0; j<newSize; j++) // for each position,

{

doAnagram(newSize-1); // anagram remaining if(newSize==2) // if innermost,

displayWord(); // display it

rotate(newSize); // rotate word

}

}

Each time the doAnagram() method calls itself, the size of the word is one letter smaller, and the starting position is one cell further to the right, as shown in Figure 6.8.

n

e

t

s

0

1

2

3

Level 4

newSize = 1

position = 3

Level 3

newSize = 2

position = 2

Level 2

newSize = 3

position = 1

Level 1

newSize = 4

position = 0

**FIGURE 6.8**

Smaller and smaller words.

Listing 6.2 shows the complete anagram.java program. The main() routine gets a word from the user, inserts it into a character array so it can be dealt with conveniently, and then calls doAnagram().

**LISTING 6.2**

The anagram.java Program

// anagram.java

// creates anagrams

// to run this program: C>java AnagramApp

import java.io.*;

////////////////////////////////////////////////////////////////

class AnagramApp

{

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**LISTING 6.2**

Continued

static int size;

static int count;

static char[] arrChar = new char[100];

public static void main(String[] args) throws IOException

{

System.out.print(“Enter a word: “); // get word

String input = getString();

size = input.length(); // find its size count = 0;

for(int j=0; j<size; j++) // put it in array arrChar[j] = input.charAt(j);

doAnagram(size); // anagram it

} // end main()

//-----------------------------------------------------------

public static void doAnagram(int newSize)

{

if(newSize == 1) // if too small, return; // go no further

for(int j=0; j<newSize; j++) // for each position,

{

doAnagram(newSize-1); // anagram remaining if(newSize==2) // if innermost,

displayWord(); // display it

rotate(newSize); // rotate word

}

}

//-----------------------------------------------------------

// rotate left all chars from position to end

public static void rotate(int newSize)

{

int j;

int position = size - newSize;

char temp = arrChar[position]; // save first letter for(j=position+1; j<size; j++) // shift others left arrChar[j-1] = arrChar[j];

arrChar[j-1] = temp; // put first on right

}

//-----------------------------------------------------------

public static void displayWord()

{

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**LISTING 6.2**

Continued

if(count < 99)

System.out.print(“ “);

if(count < 9)

System.out.print(“ “);

System.out.print(++count + “ “);

for(int j=0; j<size; j++)

System.out.print( arrChar[j] );

System.out.print(“ “);

System.out.flush();

if(count%6 == 0)

System.out.println(“”);

}

//-----------------------------------------------------------

public static String getString() throws IOException

{

String s = br.readLine();

return s;

}

//-------------------------------------------------------------

} // end class AnagramApp

////////////////////////////////////////////////////////////////

The rotate() method rotates the word one position left as described earlier. The displayWord() method displays the entire word and adds a count to make it easy to see how many words have been displayed. Here’s some sample interaction with the program:

Enter a word: cats

1 cats 2 cast 3 ctsa 4 ctas 5 csat 6 csta 7 atsc 8 atcs 9 asct 10 astc 11 acts 12 acst 13 tsca 14 tsac 15 tcas 16 tcsa 17 tasc 18 tacs 19 scat 20 scta 21 satc 22 sact 23 stca 24 stac (Is it only coincidence that

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**A Recursive Binary Search**

Remember the binary search we discussed in Chapter 2, “Arrays”? We wanted to find a given cell in an ordered array using the fewest number of comparisons. The solution was to divide the array in half, see which half the desired cell lay in, divide that half in half again, and so on. Here’s what the original find() method looked like:

//-----------------------------------------------------------

public int find(long searchKey)

{

int lowerBound = 0;

int upperBound = nElems-1;

int curIn;

while(true)

{

curIn = (lowerBound + upperBound ) / 2;

if(a[curIn]==searchKey)

return curIn; // found it

else if(lowerBound > upperBound)

return nElems; // can’t find it

else // divide range

{

if(a[curIn] < searchKey)

lowerBound = curIn + 1; // it’s in upper half

else

upperBound = curIn - 1; // it’s in lower half

} // end else divide range

} // end while

} // end find()

//-----------------------------------------------------------

You might want to reread the section on binary searches in ordered arrays in Chapter 2, which describes how this method works. Also, run the Ordered Workshop applet from that chapter if you want to see a binary search in action.

We can transform this loop-based method into a recursive method quite easily. In the loop-based method, we change lowerBound or upperBound to specify a new range and then cycle through the loop again. Each time through the loop we divide the range (roughly) in half.

**Recursion Replaces the Loop**

In the recursive approach, instead of changing lowerBound or upperBound, we call find() again with the new values of lowerBound or upperBound as arguments. The loop disappears, and its place is taken by the recursive calls. Here’s how that looks:

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private int recFind(long searchKey, int lowerBound, int upperBound)

{

int curIn;

curIn = (lowerBound + upperBound ) / 2;

if(a[curIn]==searchKey)

return curIn; // found it

else if(lowerBound > upperBound)

return nElems; // can’t find it

else // divide range

{

if(a[curIn] < searchKey) // it’s in upper half return recFind(searchKey, curIn+1, upperBound);

else // it’s in lower half

return recFind(searchKey, lowerBound, curIn-1);

} // end else divide range

} // end recFind()

The class user, represented by main(), may not know how many items are in the array when it calls find(), and in any case shouldn’t be burdened with having to know what values of upperBound and lowerBound to set initially. Therefore, we supply an intermediate public method, find(), which main() calls with only one argument, the value of the search key. The find() method supplies the proper initial values of lowerBound and upperBound (0 and nElems-1) and then calls the private, recursive method recFind(). The find() method looks like this: public int find(long searchKey)

{

return recFind(searchKey, 0, nElems-1);

}

Listing 6.3 shows the complete listing for the binarySearch.java program.

**LISTING 6.3**

The binarySearch.java Program

// binarySearch.java

// demonstrates recursive binary search

// to run this program: C>java BinarySearchApp

////////////////////////////////////////////////////////////////

class ordArray

{

private long[] a; // ref to array a

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**LISTING 6.3**

Continued

private int nElems; // number of data items

//-----------------------------------------------------------

public ordArray(int max) // constructor

{

a = new long[max]; // create array

nElems = 0;

}

//-----------------------------------------------------------

public int size()

{ return nElems; }

//-----------------------------------------------------------

public int find(long searchKey)

{

return recFind(searchKey, 0, nElems-1);

}

//-----------------------------------------------------------

private int recFind(long searchKey, int lowerBound, int upperBound)

{

int curIn;

curIn = (lowerBound + upperBound ) / 2;

if(a[curIn]==searchKey)

return curIn; // found it

else if(lowerBound > upperBound)

return nElems; // can’t find it

else // divide range

{

if(a[curIn] < searchKey) // it’s in upper half return recFind(searchKey, curIn+1, upperBound);

else // it’s in lower half

return recFind(searchKey, lowerBound, curIn-1);

} // end else divide range

} // end recFind()

//-----------------------------------------------------------

public void insert(long value) // put element into array

{

int j;

for(j=0; j<nElems; j++) // find where it goes if(a[j] > value) // (linear search)

break;

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271

**LISTING 6.3**

Continued

for(int k=nElems; k>j; k--) // move bigger ones up a[k] = a[k-1];

a[j] = value; // insert it

nElems++; // increment size

} // end insert()

//-----------------------------------------------------------

public void display() // displays array contents

{

for(int j=0; j<nElems; j++) // for each element, System.out.print(a[j] + “ “); // display it

System.out.println(“”);

}

//-----------------------------------------------------------

} // end class ordArray

////////////////////////////////////////////////////////////////

class BinarySearchApp

{

public static void main(String[] args)

{

int maxSize = 100; // array size

ordArray arr; // reference to array arr = new ordArray(maxSize); // create the array

arr.insert(72); // insert items

arr.insert(90);

arr.insert(45);

arr.insert(126);

arr.insert(54);

arr.insert(99);

arr.insert(144);

arr.insert(27);

arr.insert(135);

arr.insert(81);

arr.insert(18);

arr.insert(108);

arr.insert(9);

arr.insert(117);

arr.insert(63);

arr.insert(36);

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**LISTING 6.3**

Continued

arr.display(); // display array

int searchKey = 27; // search for item

if( arr.find(searchKey) != arr.size() )

System.out.println(“Found “ + searchKey);

else

System.out.println(“Can’t find “ + searchKey);

} // end main()

} // end class BinarySearchApp

////////////////////////////////////////////////////////////////

In main() we insert 16 items into the array. The insert() method arranges them in sorted order; they’re then displayed. Finally, we use find() to try to find the item with a key value of 27. Here’s some sample output:

9 18 27 36 45 54 63 72 81 90 99 108 117 126 135 144

Found 27

In binarySearch.java there are 16 items in an array. Figure 6.9 shows how the recFind() method in this program calls itself over and over, each time with a smaller range than before. When the innermost version of the method finds the desired item, which has the key value 27, it returns with the index value of the item, which is 2 (as can be seen in the display of ordered data). This value is then returned from each version of recFind() in turn; finally, find() returns it to the class user.

The recursive binary search has the same big O efficiency as the non-recursive version: O(logN). It is somewhat more elegant, but may be slightly slower.

**Divide-and-Conquer Algorithms**

The recursive binary search is an example of the

The divide-and-conquer approach is commonly used with recursion, although, as we saw in the binary search in Chapter 2, you can also use a non-recursive approach.

A divide-and-conquer approach usually involves a method that contains two recursive calls to itself, one for each half of the problem. In the binary search, there are two such calls, but only one of them is actually executed. (Which one depends on the value of the key.) The mergesort, which we’ll encounter later in this chapter, actually executes both recursive calls (to sort two halves of an array).

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called with

lowerBound=0

upperBound=15

lowerBound=0

upperBound=15

lowerBound=0

upperBound=6

lowerBound=0

upperBound=2

lowerBound=2

upperBound=2

Found it at 2

Return 2

Version 5

Version 4

Return 2

Version 3

Return 2

Version 2

Return 2

Version 1

Return 2

Returns 2

**FIGURE 6.9**

The recursive binarySearch() method.

**The Towers of Hanoi**

The Towers of Hanoi is an ancient puzzle consisting of a number of disks placed on three columns, as shown in Figure 6.10.

The disks all have different diameters and holes in the middle so they will fit over the columns. All the disks start out on column A. The object of the puzzle is to transfer all the disks from column A to column C. Only one disk can be moved at a time, and no disk can be placed on a disk that’s smaller than itself.

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**FIGURE 6.10**

The Towers of Hanoi.

There’s an ancient myth that somewhere in India, in a remote temple, monks labor day and night to transfer 64 golden disks from one of three diamond-studded towers to another. When they are finished, the world will end. Any alarm you may feel, however, will be dispelled when you see how long it takes to solve the puzzle for far fewer than 64 disks.

**The Towers Workshop Applet**

Start up the Towers Workshop applet. You can attempt to solve the puzzle yourself by using the mouse to drag the topmost disk to another tower. Figure 6.11 shows how the towers look after several moves have been made.

**FIGURE 6.11**

The Towers Workshop applet.

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There are three ways to use the Workshop applet:

• You can attempt to solve the puzzle manually, by dragging the disks from tower to tower.

• You can repeatedly press the Step button to watch the algorithm solve the puzzle. At each step in the solution, a message is displayed, telling you what the algorithm is doing.

• You can press the Run button and watch the algorithm solve the puzzle with no intervention on your part; the disks zip back and forth between the posts.

To restart the puzzle, type in the number of disks you want to use, from 1 to 10, and press New twice. (After the first time, you’re asked to verify that restarting is what you want to do.) The specified number of disks will be arranged on tower A. Once you drag a disk with the mouse, you can’t use Step or Run; you must start over with New. However, you can switch to manual in the middle of stepping or running, and you can switch to Step when you’re running, and Run when you’re stepping.

Try solving the puzzle manually with a small number of disks, say three or four.

Work up to higher numbers. The applet gives you the opportunity to learn intuitively how the problem is solved.

**Moving Subtrees**

Let’s call the initial tree-shaped (or pyramid-shaped) arrangement of disks on tower A a

A

B

C

1

2

4

3

**FIGURE 6.12**

A subtree on tower B.

These subtrees form many times in the solution of the puzzle. This happens because the creation of a subtree is the only way to transfer a larger disk from one tower to

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another: All the smaller disks must be placed on an intermediate tower, where they naturally form a subtree.

Here’s a rule of thumb that may help when you try to solve the puzzle manually. If the subtree you’re trying to move has an odd number of disks, start by moving the topmost disk directly to the tower where you want the subtree to go. If you’re trying to move a subtree with an even number of disks, start by moving the topmost disk to the intermediate tower.

**The Recursive Algorithm**

The solution to the Towers of Hanoi puzzle can be expressed recursively using the notion of subtrees. Suppose you want to move all the disks from a source tower (call it S) to a destination tower (call it D). You have an intermediate tower available (call it I). Assume there are n disks on tower S. Here’s the algorithm: **1. **Move the subtree consisting of the top n-1 disks from S to I.

**2. **Move the remaining (largest) disk from S to D.

**3. **Move the subtree from I to D.

When you begin, the source tower is A, the intermediate tower is B, and the destination tower is C. Figure 6.13 shows the three steps for this situation.

First, the subtree consisting of disks 1, 2, and 3 is moved to the intermediate tower B.

Then the largest disk, 4, is moved to tower C. Then the subtree is moved from B to C.

Of course, this solution doesn’t solve the problem of how to move the subtree consisting of disks 1, 2, and 3 to tower B, because you can’t move a subtree all at once; you must move it one disk at a time. Moving the three-disk subtree is not so easy. However, it’s easier than moving four disks.

As it turns out, moving three disks from A to the destination tower B can be done with the same three steps as moving four disks. That is, move the subtree consisting of the top two disks from tower A to intermediate tower C; then move disk 3 from A to B. Then move the subtree back from C to B.

How do you move a subtree of two disks from A to C? Move the subtree consisting of only one disk (1) from A to B. This is the base case: When you’re moving only one disk, you just move it; there’s nothing else to do. Then move the larger disk (2) from A to C, and replace the subtree (disk 1) on it.

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277

A

B

C

1

Move

Subtree

subtree

2

to B

3

4

a)

A

Move bottom

B

C

disk to C

1

2

4

3

b)

A

Move subtree

B

C

to C

1

2

3

4

c)

Puzzle

is

A

B

C

solved!

1

2

3

4

d)

**FIGURE 6.13**

Recursive solution to towers puzzle.

**The **towers.java **Program**

The towers.java program solves the Towers of Hanoi puzzle using this recursive approach. It communicates the moves by displaying them; this approach requires much less code than displaying the towers. It’s up to the human reading the list to actually carry out the moves.

The code is simplicity itself. The main() routine makes a single call to the recursive method doTowers(). This method then calls itself recursively until the puzzle is solved. In this version, shown in Listing 6.4, there are initially only three disks, but you can recompile the program with any number.

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**CHAPTER 6**

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**LISTING 6.4**

The towers.java Program

// towers.java

// solves the towers of Hanoi puzzle

// to run this program: C>java TowersApp

////////////////////////////////////////////////////////////////

class TowersApp

{

static int nDisks = 3;

public static void main(String[] args)

{

doTowers(nDisks, ‘A’, ‘B’, ‘C’);

}

//-----------------------------------------------------------

public static void doTowers(int topN,

char from, char inter, char to)

{

if(topN==1)

System.out.println(“Disk 1 from “ + from + “ to “+ to); else

{

doTowers(topN-1, from, to, inter); // from-->inter System.out.println(“Disk “ + topN +

“ from “ + from + “ to “+ to);

doTowers(topN-1, inter, from, to); // inter-->to

}

}

//----------------------------------------------------------

} // end class TowersApp

////////////////////////////////////////////////////////////////

Remember that three disks are moved from A to C. Here’s the output from the program:

Disk 1 from A to C

Disk 2 from A to B

Disk 1 from C to B

Disk 3 from A to C

Disk 1 from B to A

Disk 2 from B to C

Disk 1 from A to C

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The arguments to doTowers() are the number of disks to be moved, and the source (from), intermediate (inter), and destination (to) towers to be used. The number of disks decreases by 1 each time the method calls itself. The source, intermediate, and destination towers also change.

Here is the output with additional notations that show when the method is entered and when it returns, its arguments, and whether a disk is moved because it’s the base case (a subtree consisting of only one disk) or because it’s the remaining bottom disk after a subtree has been moved:

Enter (3 disks): s=A, i=B, d=C

Enter (2 disks): s=A, i=C, d=B

Enter (1 disk): s=A, i=B, d=C

Base case: move disk 1 from A to C

Return (1 disk)

Move bottom disk 2 from A to B

Enter (1 disk): s=C, i=A, d=B

Base case: move disk 1 from C to B

Return (1 disk)

Return (2 disks)

Move bottom disk 3 from A to C

Enter (2 disks): s=B, i=A, d=C

Enter (1 disk): s=B, i=C, d=A

Base case: move disk 1 from B to A

Return (1 disk)

Move bottom disk 2 from B to C

Enter (1 disk): s=A, i=B, d=C

Base case: move disk 1 from A to C

Return (1 disk)

Return (2 disks)

Return (3 disks)

If you study this output along with the source code for doTower(), it should become clear exactly how the method works. It’s amazing that such a small amount of code can solve such a seemingly complicated problem.

**mergesort**

Our final example of recursion is the mergesort. This is a much more efficient sorting technique than those we saw in Chapter 3, “Simple Sorting,” at least in terms of speed. While the bubble, insertion, and selection sorts take O(N2) time, the mergesort is O(N*logN). The graph in Figure 2.9 (in Chapter 2) shows how much faster this is. For example, if N (the number of items to be sorted) is 10,000, then N2 is

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100,000,000, while N*logN is only 40,000. If sorting this many items required 40

seconds with the mergesort, it would take almost 28 hours for the insertion sort.

The mergesort is also fairly easy to implement. It’s conceptually easier than quicksort and the Shell short, which we’ll encounter in the next chapter.

The downside of the mergesort is that it requires an additional array in memory, equal in size to the one being sorted. If your original array barely fits in memory, the mergesort won’t work. However, if you have enough space, it’s a good choice.

**Merging Two Sorted Arrays**

The heart of the mergesort algorithm is the merging of two already-sorted arrays.

Merging two sorted arrays A and B creates a third array, C, that contains all the elements of A and B, also arranged in sorted order. We’ll examine the merging process first; later we’ll see how it’s used in sorting.

Imagine two sorted arrays. They don’t need to be the same size. Let’s say array A has 4 elements and array B has 6. They will be merged into an array C that starts with 10

empty cells. Figure 6.14 shows these arrays.

A

23 47 81 95

0

1

2

3

❸

❺

❾

❿

C

0

1

2

3

4

5

6

7

8

9

❶

❷

❹

❻ ❼

❽

B

7

14 39 55 62 74

0

1

2

3

4

5

a) Before Merge

C

7

14 23 39 47 55 62 74 81

95

0

1

2

3

4

5

6

7

8

9

b) After Merge

**FIGURE 6.14**

Merging two arrays.

In the figure, the circled numbers indicate the order in which elements are transferred from A and B to C. Table 6.3 shows the comparisons necessary to determine which element will be copied. The steps in the table correspond to the steps in the figure. Following each comparison, the smaller element is copied to A.

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**TABLE 6.3**

Merging Operations

**Step**

**Comparison (If Any)**

**Copy**

1

Compare 23 and 7

Copy 7 from B to C

2

Compare 23 and 14

Copy 14 from B to C

3

Compare 23 and 39

Copy 23 from A to C

4

Compare 39 and 47

Copy 39 from B to C

5

Compare 55 and 47

Copy 47 from A to C

6

Compare 55 and 81

Copy 55 from B to C

7

Compare 62 and 81

Copy 62 from B to C

8

Compare 74 and 81

Copy 74 from B to C

9

Copy 81 from A to C

10

Copy 95 from A to C

Notice that, because B is empty following step 8, no more comparisons are necessary; all the remaining elements are simply copied from A into C.

Listing 6.5 shows a Java program that carries out the merge shown in Figure 6.14

and Table 6.3. This is not a recursive program; it is a prelude to understanding mergesort.

**LISTING 6.5**

The merge.java Program

// merge.java

// demonstrates merging two arrays into a third

// to run this program: C>java MergeApp

////////////////////////////////////////////////////////////////

class MergeApp

{

public static void main(String[] args)

{

int[] arrayA = {23, 47, 81, 95};

int[] arrayB = {7, 14, 39, 55, 62, 74};

int[] arrayC = new int[10];

merge(arrayA, 4, arrayB, 6, arrayC);

display(arrayC, 10);

} // end main()

//-----------------------------------------------------------

// merge A and B into C

public static void merge( int[] arrayA, int sizeA,

int[] arrayB, int sizeB,

int[] arrayC )

{

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**LISTING 6.5**

Continued

int aDex=0, bDex=0, cDex=0;

while(aDex < sizeA && bDex < sizeB) // neither array empty if( arrayA[aDex] < arrayB[bDex] )

arrayC[cDex++] = arrayA[aDex++];

else

arrayC[cDex++] = arrayB[bDex++];

while(aDex < sizeA) // arrayB is empty, arrayC[cDex++] = arrayA[aDex++]; // but arrayA isn’t while(bDex < sizeB) // arrayA is empty, arrayC[cDex++] = arrayB[bDex++]; // but arrayB isn’t

} // end merge()

//-----------------------------------------------------------

// display array

public static void display(int[] theArray, int size)

{

for(int j=0; j<size; j++)

System.out.print(theArray[j] + “ “);

System.out.println(“”);

}

//-----------------------------------------------------------

} // end class MergeApp

////////////////////////////////////////////////////////////////

In main() the arrays arrayA, arrayB, and arrayC are created; then the merge() method is called to merge arrayA and arrayB into arrayC, and the resulting contents of arrayC

are displayed. Here’s the output:

7 14 23 39 47 55 62 74 81 95

The merge() method has three while loops. The first steps along both arrayA and arrayB, comparing elements and copying the smaller of the two into arrayC.

The second while loop deals with the situation when all the elements have been transferred out of arrayB, but arrayA still has remaining elements. (This is what happens in the example, where 81 and 95 remain in arrayA.) The loop simply copies the remaining elements from arrayA into arrayC.

The third loop handles the similar situation when all the elements have been transferred out of arrayA, but arrayB still has remaining elements; they are copied to arrayC.

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**Sorting by Merging**

The idea in the mergesort is to divide an array in half, sort each half, and then use the merge() method to merge the two halves into a single sorted array. How do you sort each half? This chapter is about recursion, so you probably already know the answer: You divide the half into two quarters, sort each of the quarters, and merge them to make a sorted half.

Similarly, each pair of 8ths is merged to make a sorted quarter, each pair of 16ths is merged to make a sorted 8th, and so on. You divide the array again and again until you reach a subarray with only one element. This is the base case; it’s assumed an array with one element is already sorted.

We’ve seen that generally something is reduced in size each time a recursive method calls itself, and built back up again each time the method returns. In mergeSort() the range is divided in half each time this method calls itself, and each time it returns it merges two smaller ranges into a larger one.

As mergeSort() returns from finding two arrays of one element each, it merges them into a sorted array of two elements. Each pair of resulting 2-element arrays is then merged into a 4-element array. This process continues with larger and larger arrays until the entire array is sorted. This is easiest to see when the original array size is a power of 2, as shown in Figure 6.15.

First, in the bottom half of the array, range 0-0 and range 1-1 are merged into range 0-1. Of course, 0-0 and 1-1 aren’t really ranges; they’re only one element, so they are base cases. Similarly, 2-2 and 3-3 are merged into 2-3. Then ranges 0-1 and 2-3 are merged into 0-3.

In the top half of the array, 4-4 and 5-5 are merged into 4-5, 6-6 and 7-7 are merged into 6-7, and 4-5 and 6-7 are merged into 4-7. Finally, the top half, 0-3, and the bottom half, 4-7, are merged into the complete array, 0-7, which is now sorted.

When the array size is not a power of 2, arrays of different sizes must be merged. For example, Figure 6.16 shows the situation when the array size is 12. Here an array of size 2 must be merged with an array of size 1 to form an array of size 3.

First, the 1-element ranges 0-0 and 1-1 are merged into the 2-element range 0-1.

Then range 0-1 is merged with the 1-element range 2-2. This creates a 3-element range 0-2. It’s merged with the 3-element range 3-5. The process continues until the array is sorted.

Notice that in mergesort we don’t merge two separate arrays into a third one, as we demonstrated in the merge.java program. Instead, we merge parts of a single array into itself.

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0

1

2

3

4

5

6

7

64

21

33

70

12

85

44

3

❶ 21

64

33

70

❷

❸ 21

33

64

70

12

85

❹

❺

3

44

❻

3

12

44

85

3

12

21

33

44

64

70

85

❼

**FIGURE 6.15**

Merging larger and larger arrays.

You may wonder where all these subarrays are located in memory. In the algorithm, a workspace array of the same size as the original array is created. The subarrays are stored in sections of the workspace array. This means that subarrays in the original array are copied to appropriate places in the workspace array. After each merge, the workspace array is copied back into the original array.

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285

0

1

2

3

4

5

6

7

8

9

10

11

64

21

33

70

12

85

44

3

97

24

51

40

❶

21

64

❷

21

33

64

12

70

❸

12

70

85

❹

❺

3

44

❻

3

44

97

❼

24

51

24

40

51

❽

❾

12

21

33

64

70

85

❿

3

24

40

44

51

97

3

12

21

24

33

40

44

51

64

70

85

97

**FIGURE 6.16**

Array size not a power of 2.

**The MergeSort Workshop Applet**

This sorting process is easier to appreciate when you see it happening before your very eyes. Start up the MergeSort Workshop applet. Repeatedly pressing the Step

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button will execute mergesort step by step. Figure 6.17 shows what it looks like after the first three presses.

**FIGURE 6.17**

The MergeSort Workshop applet.

The Lower and Upper arrows show the range currently being considered by the algorithm, and the Mid arrow shows the middle part of the range. The range starts as the entire array and then is halved each time the mergeSort() method calls itself. When the range is one element, mergeSort() returns immediately; that’s the base case.

Otherwise, the two subarrays are merged. The applet provides messages, such as Entering mergeSort: 0-5, to tell you what it’s doing and the range it’s operating on.

Many steps involve the mergeSort() method calling itself or returning. Comparisons and copies are performed only during the merge process, when you’ll see messages like Merged 0-0 and 1-1 into workspace. You can’t see the merge happening because the workspace isn’t shown. However, you can see the result when the appropriate section of the workspace is copied back into the original (visible) array: The bars in the specified range will appear in sorted order.

First, the first two bars will be sorted, then the first three bars, then the two bars in the range 3-4, then the three bars in the range 3-5, then the six bars in the range 0-5, and so on, corresponding to the sequence shown in Figure 6.16. Eventually, all the bars will be sorted.

You can cause the algorithm to run continuously by pressing the Run button. You can stop this process at any time by pressing Step, single-step as many times as you want, and resume running by pressing Run again.

As in the other sorting Workshop applets, pressing New resets the array with a new group of unsorted bars, and toggles between random and inverse arrangements. The Size button toggles between 12 bars and 100 bars.

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Watching the algorithm run with 100 inversely sorted bars is especially instructive.

The resulting patterns show clearly how each range is sorted individually and merged with its other half, and how the ranges grow larger and larger.

**The **mergeSort.java **Program**

In a moment we’ll look at the entire mergeSort.java program. First, let’s focus on the method that carries out the mergesort. Here it is:

private void recMergeSort(long[] workSpace, int lowerBound, int upperBound)

{

if(lowerBound == upperBound) // if range is 1, return; // no use sorting else

{ // find midpoint int mid = (lowerBound+upperBound) / 2;

// sort low half

recMergeSort(workSpace, lowerBound, mid);

// sort high half

recMergeSort(workSpace, mid+1, upperBound);

// merge them

merge(workSpace, lowerBound, mid+1, upperBound);

} // end else

} // end recMergeSort

As you can see, besides the base case, there are only four statements in this method.

One computes the midpoint, there are two recursive calls to recMergeSort() (one for each half of the array), and finally a call to merge() to merge the two sorted halves.

The base case occurs when the range contains only one element (lowerBound==upperBound) and results in an immediate return.

In the mergeSort.java program, the mergeSort() method is the one actually seen by the class user. It creates the array workSpace[] and then calls the recursive routine recMergeSort() to carry out the sort. The creation of the workspace array is handled in mergeSort() because doing it in recMergeSort() would cause the array to be created anew with each recursive call, an inefficiency.

The merge() method in the previous merge.java program (Listing 6.5) operated on three separate arrays: two source arrays and a destination array. The merge() routine in the mergeSort.java program operates on a single array: the theArray member of the DArray class. The arguments to this merge() method are the starting point of the low-half subarray, the starting point of the high-half subarray, and the upper bound of the high-half subarray. The method calculates the sizes of the subarrays based on this information.

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Listing 6.6 shows the complete mergeSort.java program, which uses a variant of the array classes from Chapter 2, adding the mergeSort() and recMergeSort() methods to the DArray class. The main() routine creates an array, inserts 12 items, displays the array, sorts the items with mergeSort(), and displays the array again.

**LISTING 6.6**

The mergeSort.java Program

// mergeSort.java

// demonstrates recursive merge sort

// to run this program: C>java MergeSortApp

////////////////////////////////////////////////////////////////

class DArray

{

private long[] theArray; // ref to array theArray private int nElems; // number of data items

//-----------------------------------------------------------

public DArray(int max) // constructor

{

theArray = new long[max]; // create array

nElems = 0;

}

//-----------------------------------------------------------

public void insert(long value) // put element into array

{

theArray[nElems] = value; // insert it

nElems++; // increment size

}

//-----------------------------------------------------------

public void display() // displays array contents

{

for(int j=0; j<nElems; j++) // for each element, System.out.print(theArray[j] + “ “); // display it System.out.println(“”);

}

//-----------------------------------------------------------

public void mergeSort() // called by main()

{ // provides workspace long[] workSpace = new long[nElems];

recMergeSort(workSpace, 0, nElems-1);

}

//-----------------------------------------------------------

private void recMergeSort(long[] workSpace, int lowerBound, int upperBound)

{

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**LISTING 6.6**

Continued

if(lowerBound == upperBound) // if range is 1, return; // no use sorting else

{ // find midpoint int mid = (lowerBound+upperBound) / 2;

// sort low half

recMergeSort(workSpace, lowerBound, mid);

// sort high half

recMergeSort(workSpace, mid+1, upperBound);

// merge them

merge(workSpace, lowerBound, mid+1, upperBound);

} // end else

} // end recMergeSort()

//-----------------------------------------------------------

private void merge(long[] workSpace, int lowPtr,

int highPtr, int upperBound)

{

int j = 0; // workspace index int lowerBound = lowPtr;

int mid = highPtr-1;

int n = upperBound-lowerBound+1; // # of items while(lowPtr <= mid && highPtr <= upperBound) if( theArray[lowPtr] < theArray[highPtr] )

workSpace[j++] = theArray[lowPtr++];

else

workSpace[j++] = theArray[highPtr++];

while(lowPtr <= mid)

workSpace[j++] = theArray[lowPtr++];

while(highPtr <= upperBound)

workSpace[j++] = theArray[highPtr++];

for(j=0; j<n; j++)

theArray[lowerBound+j] = workSpace[j];

} // end merge()

//-----------------------------------------------------------

} // end class DArray

////////////////////////////////////////////////////////////////

class MergeSortApp

{

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**LISTING 6.6**

Continued

public static void main(String[] args)

{

int maxSize = 100; // array size

DArray arr; // reference to array arr = new DArray(maxSize); // create the array

arr.insert(64); // insert items

arr.insert(21);

arr.insert(33);

arr.insert(70);

arr.insert(12);

arr.insert(85);

arr.insert(44);

arr.insert(3);

arr.insert(99);

arr.insert(0);

arr.insert(108);

arr.insert(36);

arr.display(); // display items

arr.mergeSort(); // merge sort the array arr.display(); // display items again

} // end main()

} // end class MergeSortApp

////////////////////////////////////////////////////////////////

The output from the program is simply the display of the unsorted and sorted arrays: 64 21 33 70 12 85 44 3 99 0 108 36

0 3 12 21 33 36 44 64 70 85 99 108

If we put additional statements in the recMergeSort() method, we could generate a running commentary on what the program does during a sort. The following output shows how this might look for the four-item array {64, 21, 33, 70}. (You can think of this as the lower half of the array in Figure 6.15.) Entering 0-3

Will sort low half of 0-3

Entering 0-1

Will sort low half of 0-1

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Entering 0-0

Base-Case Return 0-0

Will sort high half of 0-1

Entering 1-1

Base-Case Return 1-1

Will merge halves into 0-1

Return 0-1 theArray=21 64 33 70

Will sort high half of 0-3

Entering 2-3

Will sort low half of 2-3

Entering 2-2

Base-Case Return 2-2

Will sort high half of 2-3

Entering 3-3

Base-Case Return 3-3

Will merge halves into 2-3

Return 2-3 theArray=21 64 33 70

Will merge halves into 0-3

Return 0-3 theArray=21 33 64 70

This is roughly the same content as would be generated by the MergeSort Workshop applet if it could sort four items. Study of this output, and comparison with the code for recMergeSort() and Figure 6.15, will reveal the details of the sorting process.

**Efficiency of the mergesort**

As we noted, the mergesort runs in O(N*logN) time. How do we know this? Let’s see how we can figure out the number of times a data item must be copied and the number times it must be compared with another data item during the course of the algorithm. We assume that copying and comparing are the most time-consuming operations; that the recursive calls and returns don’t add much overhead.

**Number of Copies**

Consider Figure 6.15. Each cell below the top line represents an element copied from the array into the workspace.

Adding up all the cells in Figure 6.15 (the seven numbered steps) shows there are 24

copies necessary to sort 8 items. Log 8 is 3, so 8*log 8 equals 24. This shows that, for 2

2

the case of 8 items, the number of copies is proportional to N*log N.

2

Another way to look at this calculation is that, to sort 8 items requires 3

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In Figure 6.15, by considering only half the graph, you can see that 8 copies are necessary for an array of 4 items (steps 1, 2, and 3), and 2 copies are necessary for 2

items. Similar calculations provide the number of copies necessary for larger arrays.

Table 6.4 summarizes this information.

**TABLE 6.4**

Number of Operations When N Is a Power of 2

**Number of**

**Copies into**

**Workspace**

**Comparisons**

**N**

**log N**

**(N*log N)**

**Total Copies**

**Max (Min)**

**2**

**2**

2

1

2

4

1 (1)

4

2

8

16

5 (4)

8

3

24

48

17 (12)

16

4

64

128

49 (32)

32

5

160

320

129 (80)

64

6

384

768

321 (192)

128

7

896

1792

769 (448)

Actually, the items are not only copied into the workspace, they’re also copied back into the original array. This doubles the number of copies, as shown in the Total Copies column. The final column of Table 6.4 shows comparisons, which we’ll return to in a moment.

It’s harder to calculate the number of copies and comparisons when N is not a multiple of 2, but these numbers fall between those that are a power of 2. For 12 items, there are 88 total copies, and for 100 items, 1,344 total copies.

**Number of Comparisons**

In the mergesort algorithm, the number of comparisons is always somewhat less than the number of copies. How much less? Assuming the number of items is a power of 2, for each individual merging operation, the maximum number of comparisons is always one less than the number of items being merged, and the minimum is half the number of items being merged. You can see why this is true in Figure 6.18, which shows two possibilities when trying to merge two arrays of four items each.

In the first case, the items interleave, and seven comparisons must be made to merge them. In the second case, all the items in one array are smaller than all the items in the other, so only four comparisons need be made.

There are many merges for each sort, so we must add the comparisons for each one.

Referring back to Figure 6.15, you can see that seven merge operations are required to sort eight items. The number of items being merged and the resulting number of comparisons are shown in Table 6.5.

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A

24 47 63 85

Comparisons

❷

❹

❻

❽

1. 12-24

2. 24-34

3. 34-47

C

4. 47-55

5. 55-63

❶

❸

❺

❼

6. 63-79

7. 79-85

B

12 34 55 79

a) Worst-case Scenario

12 24 34 47

Comparisons

❶ ❷ ❸ ❹

1. 12-55

2. 24-55

3. 34-55

4. 47-55

❺ ❻ ❼ ❽

55 63 79 85

b) Best-case Scenario

**FIGURE 6.18**

Maximum and minimum comparisons.

**TABLE 6.5**

Comparisons Involved in Sorting 8 Items

**Step Number**

**1**

**2**

**3**

**4**

**5**

**6**

**7**

**Totals**

Number of items

2

2

4

2

2

4

8

24

being merged

(N)

Maximum

1

1

3

1

1

3

7

17

comparisons

(N-1)

Minimum

1

1

2

1

1

2

4

12

comparisons

(N/2)

For each merge, the maximum number of comparisons is one less than the number of items. Adding these figures for all the merges gives us a total of 17.

The minimum number of comparisons is always half the number of items being merged, and adding these figures for all the merges results in 12 comparisons.

Similar arithmetic results in the Comparisons columns for Table 6.4. The actual

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number of comparisons to sort a specific array depends on how the data is arranged, but it will be somewhere between the maximum and minimum values.

**Eliminating Recursion**

Some algorithms lend themselves to a recursive approach, some don’t. As we’ve seen, the recursive triangle() and factorial() methods can be implemented more efficiently using a simple loop. However, various divide-and-conquer algorithms, such as mergesort, work very well as recursive routines.

Often an algorithm is easy to conceptualize as a recursive method, but in practice the recursive approach proves to be inefficient. In such cases, it’s useful to transform the recursive approach into a non-recursive approach. Such a transformation can often make use of a stack.

**Recursion and Stacks**

There is a close relationship between recursion and stacks. In fact, most compilers implement recursion by using stacks. As we noted, when a method is called, the compiler pushes the arguments to the method and the return address (where control will go when the method returns) on the stack, and then transfers control to the method. When the method returns, it pops these values off the stack. The arguments disappear, and control returns to the return address.

**Simulating a Recursive Method**

In this section we’ll demonstrate how any recursive solution can be transformed into a stack-based solution. Remember the recursive triangle() method from the first section in this chapter? Here it is again:

int triangle(int n)

{

if(n==1)

return 1;

else

return( n + triangle(n-1) );

}

We’re going to break this algorithm down into its individual operations, making each operation one case in a switch statement. (You can perform a similar decomposition using goto statements in C++ and some other languages, but Java doesn’t support goto.)

The switch statement is enclosed in a method called step(). Each call to step() causes one case section within the switch to be executed. Calling step() repeatedly will eventually execute all the code in the algorithm.

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The triangle() method we just saw performs two kinds of operations. First, it carries out the arithmetic necessary to compute triangular numbers. This involves checking if n is 1, and adding n to the results of previous recursive calls. However, triangle() also performs the operations necessary to manage the method itself, including transfer of control, argument access, and the return address. These operations are not visible by looking at the code; they’re built into all methods. Here, roughly speaking, is what happens during a call to a method:

• When a method is called, its arguments and the return address are pushed onto a stack.

• A method can access its arguments by peeking at the top of the stack.

• When a method is about to return, it peeks at the stack to obtain the return address, and then pops both this address and its arguments off the stack and discards them.

The stackTriangle.java program contains three classes: Params, StackX, and StackTriangleApp. The Params class encapsulates the return address and the method’s argument, n; objects of this class are pushed onto the stack. The StackX class is similar to those in other chapters, except that it holds objects of class Params. The StackTriangleApp class contains four methods: main(), recTriangle(), step(), and the usual getInt() method for numerical input.

The main() routine asks the user for a number, calls the recTriangle() method to calculate the triangular number corresponding to n, and displays the result.

The recTriangle() method creates a StackX object and initializes codePart to 1. It then settles into a while loop, where it repeatedly calls step(). It won’t exit from the loop until step() returns true by reaching case 6, its exit point. The step() method is basically a large switch statement in which each case corresponds to a section of code in the original triangle() method. Listing 6.7 shows the stackTriangle.java program.

**LISTING 6.7**

The stackTriangle.java Program

// stackTriangle.java

// evaluates triangular numbers, stack replaces recursion

// to run this program: C>java StackTriangleApp

import java.io.*; // for I/O

////////////////////////////////////////////////////////////////

class Params // parameters to save on stack

{

public int n;

public int returnAddress;

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**LISTING 6.7**

Continued

public Params(int nn, int ra)

{

n=nn;

returnAddress=ra;

}

} // end class Params

////////////////////////////////////////////////////////////////

class StackX

{

private int maxSize; // size of StackX array private Params[] stackArray;

private int top; // top of stack

//--------------------------------------------------------------

public StackX(int s) // constructor

{

maxSize = s; // set array size

stackArray = new Params[maxSize]; // create array

top = -1; // no items yet

}

//--------------------------------------------------------------

public void push(Params p) // put item on top of stack

{

stackArray[++top] = p; // increment top, insert item

}

//--------------------------------------------------------------

public Params pop() // take item from top of stack

{

return stackArray[top--]; // access item, decrement top

}

//--------------------------------------------------------------

public Params peek() // peek at top of stack

{

return stackArray[top];

}

//--------------------------------------------------------------

} // end class StackX

////////////////////////////////////////////////////////////////

class StackTriangleApp

{

static int theNumber;

static int theAnswer;

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**LISTING 6.7**

Continued

static StackX theStack;

static int codePart;

static Params theseParams;

//-------------------------------------------------------------

public static void main(String[] args) throws IOException

{

System.out.print(“Enter a number: “);

theNumber = getInt();

recTriangle();

System.out.println(“Triangle=”+theAnswer);

} // end main()

//-------------------------------------------------------------

public static void recTriangle()

{

theStack = new StackX(10000);

codePart = 1;

while( step() == false) // call step() until it’s true

; // null statement

}

//-------------------------------------------------------------

public static boolean step()

{

switch(codePart)

{

case 1: // initial call theseParams = new Params(theNumber, 6);

theStack.push(theseParams);

codePart = 2;

break;

case 2: // method entry theseParams = theStack.peek();

if(theseParams.n == 1) // test

{

theAnswer = 1;

codePart = 5; // exit

}

else

codePart = 3; // recursive call

break;

case 3: // method call Params newParams = new Params(theseParams.n - 1, 4);

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**LISTING 6.7**

Continued

theStack.push(newParams);

codePart = 2; // go enter method

break;

case 4: // calculation theseParams = theStack.peek();

theAnswer = theAnswer + theseParams.n;

codePart = 5;

break;

case 5: // method exit theseParams = theStack.peek();

codePart = theseParams.returnAddress; // (4 or 6)

theStack.pop();

break;

case 6: // return point return true;

} // end switch

return false;

} // end triangle

//-------------------------------------------------------------

public static String getString() throws IOException

{

String s = br.readLine();

return s;

}

//-------------------------------------------------------------

public static int getInt() throws IOException

{

String s = getString();

return Integer.parseInt(s);

}

//--------------------------------------------------------------

} // end class StackTriangleApp

////////////////////////////////////////////////////////////////

This program calculates triangular numbers, just as the triangle.java program (Listing 6.1) at the beginning of the chapter did. Here’s some sample output: Enter a number: 100

Triangle=5050

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Figure 6.19 shows how the sections of code in each case relate to the various parts of the algorithm.

Case 1:

Initial call

simMeth()

Simulated method

Case 2:

Entry

Test

Done

Not Done

Case 3:

Call method

Case 4:

Do calculation

Case 5:

Exit

Case 6:

Return

point

**FIGURE 6.19**

The cases and the step() method.

The program simulates a method, but it has no name in the listing because it isn’t a real Java method. Let’s call this simulated method simMeth(). The initial call to simMeth() (at case 1) pushes the value entered by the user and a return value of 6

onto the stack and moves to the entry point of simMeth() (case 2).

At its entry (case 2), simMeth() tests whether its argument is 1. It accesses the argument by peeking at the top of the stack. If the argument is 1, this is the base case

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and control goes to simMeth()’s exit (case 5). If not, it calls itself recursively (case 3).

This recursive call consists of pushing n-1 and a return address of 4 onto the stack, and going to the method entry at case 2.

On the return from the recursive call, simMeth() adds its argument n to the value returned from the call. Finally, it exits (case 5). When it exits, it pops the last Params object off the stack; this information is no longer needed.

The return address given in the initial call was 6, so case 6 is the place where control goes when the method returns. This code returns true to let the while loop in recTriangle() know that the loop is over.

Note that in this description of simMeth()’s operation we use terms like

If you inserted some output statements in each case to see what simMeth() was doing, you could arrange for output like this:

Enter a number: 4

case 1. theAnswer=0 Stack:

case 2. theAnswer=0 Stack: (4, 6)

case 3. theAnswer=0 Stack: (4, 6)

case 2. theAnswer=0 Stack: (4, 6) (3, 4)

case 3. theAnswer=0 Stack: (4, 6) (3, 4)

case 2. theAnswer=0 Stack: (4, 6) (3, 4) (2, 4)

case 3. theAnswer=0 Stack: (4, 6) (3, 4) (2, 4)

case 2. theAnswer=0 Stack: (4, 6) (3, 4) (2, 4) (1, 4) case 5. theAnswer=1 Stack: (4, 6) (3, 4) (2, 4) (1, 4) case 4. theAnswer=1 Stack: (4, 6) (3, 4) (2, 4)

case 5. theAnswer=3 Stack: (4, 6) (3, 4) (2, 4)

case 4. theAnswer=3 Stack: (4, 6) (3, 4)

case 5. theAnswer=6 Stack: (4, 6) (3, 4)

case 4. theAnswer=6 Stack: (4, 6)

case 5. theAnswer=10 Stack: (4, 6)

case 6. theAnswer=10 Stack:

Triangle=10

The case number shows what section of code is being executed. The contents of the stack (consisting of Params objects containing n followed by a return address) are also shown. The simMeth() method is entered four times (case 2) and returns four times (case 5). Only when it starts returning does theAnswer begin to accumulate the results of the calculations.

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**What Does This Prove? **

In stackTriangle.java (Listing 6.7) we have a program that more or less systematically transforms a program that uses recursion into a program that uses a stack. This suggests that such a transformation is possible for any program that uses recursion, and in fact this is the case.

With some additional work, you can systematically refine the code we show here, simplifying it and even eliminating the switch statement entirely to make the code more efficient.

In practice, however, it’s usually more practical to rethink the algorithm from the beginning, using a stack-based approach instead of a recursive approach. Listing 6.8

shows what happens when we do that with the triangle() method.

**LISTING 6.8**

The stackTriangle2.java Program

// stackTriangle2.java

// evaluates triangular numbers, stack replaces recursion

// to run this program: C>java StackTriangle2App import java.io.*; // for I/O

////////////////////////////////////////////////////////////////

class StackX

{

private int maxSize; // size of stack array

private int[] stackArray;

private int top; // top of stack

//--------------------------------------------------------------

public StackX(int s) // constructor

{

maxSize = s;

stackArray = new int[maxSize];

top = -1;

}

//--------------------------------------------------------------

public void push(int p) // put item on top of stack

{ stackArray[++top] = p; }

//--------------------------------------------------------------

public int pop() // take item from top of stack

{ return stackArray[top--]; }

//--------------------------------------------------------------

public int peek() // peek at top of stack

{ return stackArray[top]; }

//--------------------------------------------------------------

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**LISTING 6.8**

Continued

public boolean isEmpty() // true if stack is empty

{ return (top == -1); }

//--------------------------------------------------------------

} // end class StackX

////////////////////////////////////////////////////////////////

class StackTriangle2App

{

static int theNumber;

static int theAnswer;

static StackX theStack;

public static void main(String[] args) throws IOException

{

System.out.print(“Enter a number: “);

theNumber = getInt();

stackTriangle();

System.out.println(“Triangle=”+theAnswer);

} // end main()

//-------------------------------------------------------------

public static void stackTriangle()

{

theStack = new StackX(10000); // make a stack

theAnswer = 0; // initialize answer while(theNumber > 0) // until n is 1,

{

theStack.push(theNumber); // push value

--theNumber; // decrement value

}

while( !theStack.isEmpty() ) // until stack empty,

{

int newN = theStack.pop(); // pop value,

theAnswer += newN; // add to answer

}

}

//-------------------------------------------------------------

public static String getString() throws IOException

{

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**LISTING 6.8**

Continued

String s = br.readLine();

return s;

}

//-------------------------------------------------------------

public static int getInt() throws IOException

{

String s = getString();

return Integer.parseInt(s);

}

//-------------------------------------------------------------

} // end class StackTriangle2App

Here two short while loops in the stackTriangle() method substitute for the entire step() method of the stackTriangle.java program. Of course, in this program you can see by inspection that you can eliminate the stack entirely and use a simple loop. However, in more complicated algorithms the stack must remain.

Often you’ll need to experiment to see whether a recursive method, a stack-based approach, or a simple loop is the most efficient (or practical) way to handle a particular situation.

**Some Interesting Recursive Applications**

Let’s look briefly at some other situations in which recursion is useful. You will see from the diversity of these examples that recursion can pop up in unexpected places.

We’ll examine three problems: raising a number to a power, fitting items into a knapsack, and choosing members of a mountain-climbing team. We’ll explain the concepts and leave the implementations as exercises.

**Raising a Number to a Power**

The more sophisticated pocket calculators allow you to raise a number to an arbitrary power. They usually have a key labeled something like x^y, where the circum-flex indicates that x is raised to the y power. How would you do this calculation if your calculator lacked this key? You might assume you would need to multiply x by itself y times. That is, if x was 2 and y was 8 (28), you would carry out the arithmetic for 2*2*2*2*2*2*2*2. However, for large values of y, this approach might prove tedious. Is there a quicker way?

One solution is to rearrange the problem so you multiply by multiples of 2 whenever possible, instead of by 2. Take 28 as an example. Eventually, we must involve eight 2s in the multiplication process. Let’s say we start with 2*2=4. We’ve used up two of the

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2s, but there are still six to go. However, we now have a new number to work with: 4. So we try 4*4=16. This uses four 2s (because each 4 is two 2s multiplied together).

We need to use up four more 2s, but now we have 16 to work with, and 16*16=256

uses exactly eight 2s (because each 16 has four 2s).

So we’ve found the answer to 28 with only three multiplications instead of seven.

That’s O(log N) time instead O(N).

Can we make this process into an algorithm that a computer can execute? The scheme is based on the mathematical equality xy = (x2)y/2. In our example, 28 = (22)8/2, or 28 = (22)4. This is true because raising a power to another power is the same as multiplying the powers.

However, we’re assuming our computer can’t raise a number to a power, so we can’t handle (22)4. Let’s see if we can transform this into an expression that involves only multiplication. The trick is to start by substituting a new variable for 22.

Let’s say that 22=a. Then 28 equals (22)4, which is a4. However, according to the original equality, a4 can be written (a2)2, so 28 = (a2)2.

Again we substitute a new variable for a2, say a2=c, then (c)2 can be written (c2)1, which also equals 28.

Now we have a problem we can handle with simple multiplication: c times c.

You can imbed this scheme in a recursive method—let’s call it power()—for calculating powers. The arguments are x and y, and the method returns xy. We don’t need to worry about variables like a and c anymore because x and y get new values each time the method calls itself. Its arguments are x*x and y/2. For the x=2 and y=8, the sequence of arguments and return values would be

x=2, y=8

x=4, y=4

x=16, y=2

x=256, y=1

Returning 256, x=256, y=1

Returning 256, x=16, y=2

Returning 256, x=4, y=4

Returning 256, x=2, y=8

When y is 1, we return. The answer, 256, is passed unchanged back up the sequence of methods.

We’ve shown an example in which y is an even number throughout the entire sequence of divisions. This will not usually be the case. Here’s how to revise the algorithm to deal with the situation where y is odd. Use integer division on the way down and don’t worry about a remainder when dividing y by 2. However, during the

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return process, whenever y is an odd number, do an additional multiplication by x.

Here’s the sequence for 318:

x=3, y=18

x=9, y=9

x=81, y=4

x=6561, y=2

x=43046721, y=1

Returning 43046721, x=43046721, y=1

Returning 43046721, x=6561, y=2

Returning 43046721, x=81, y=4

Returning 387420489, x=9, y=9 // y is odd; so multiply by x Returning 387420489, x=3, y=18

**The Knapsack Problem**

The Knapsack Problem is a classic in computer science. In its simplest form it involves trying to fit items of different weights into a knapsack so that the knapsack ends up with a specified total weight. You don’t need to fit in all the items.

For example, suppose you want your knapsack to weigh exactly 20 pounds, and you have five items, with weights of 11, 8, 7, 6, and 5 pounds. For small numbers of items, humans are pretty good at solving this problem by inspection. So you can probably figure out that only the 8, 7, and 5 combination of items adds up to 20.

If we want a computer to solve this problem, we’ll need to give it more detailed instructions. Here’s the algorithm:

**1. **If at any point in this process the sum of the items you selected adds up to the target, you’re done.

**2. **Start by selecting the first item. The remaining items must add up to the knapsack’s target weight minus the first item; this is a new target weight.

**3. **Try, one by one, each of the possible combinations of the remaining items.

Notice, however, that you don’t really need to try all the combinations, because whenever the sum of the items is more than the target weight, you can stop adding items.

**4. **If none of the combinations work, discard the first item, and start the whole process again with the second item.

**5. **Continue this with the third item and so on until you’ve tried all the combinations, at which point you know there is no solution.

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In the example just described, start with 11. Now we want the remaining items to add up to 9 (20 minus 11). Of these, we start with 8, which is too small. Now we want the remaining items to add up to 1 (9 minus 8). We start with 7, but that’s bigger than 1, so we try 6 and then 5, which are also too big. We’ve run out of items, so we know that any combination that includes 8 won’t add up to 9. Next we try 7, so now we’re looking for a target of 2 (9 minus 7). We continue in the same way, as summarized here:

Items: 11, 8, 7, 6, 5

==========================================

11 // Target = 20, 11 is too small

11, 8 // Target = 9, 8 is too small

11, 8, 7 // Target = 1, 7 is too big

11, 8, 6 // Target = 1, 6 is too big

11, 8, 5 // Target = 1, 5 is too big. No more items 11, 7 // Target = 9, 7 is too small

11, 7, 6 // Target = 2, 6 is too big

11, 7, 5 // Target = 2, 5 is too big. No more items 11, 6 // Target = 9, 6 is too small

11, 6, 5 // Target = 3, 5 is too big. No more items 11, 5 // Target = 9, 5 is too small. No more items 8, // Target = 20, 8 is too small

8, 7 // Target = 12, 7 is too small

8, 7, 6 // Target = 5, 6 is too big

8, 7, 5 // Target = 5, 5 is just right. Success!

As you may recognize, a recursive routine can pick the first item, and, if the item is smaller than the target, the routine can call itself with a new target to investigate the sums of all the remaining items.

**Combinations: Picking a Team**

In mathematics, a

ABC, ABD, ABE, ACD, ACE, ADE, BCD, BCE, BDE, CDE

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How would you write such a program? It turns out there’s an elegant recursive solution. It involves dividing these combinations into two groups: those that begin with A and those that don’t. Suppose we abbreviate the idea of 3 people selected from a group of 5 as (5,3). Let’s say n is the size of the group and k is the size of a team. A theorem says that

(n, k) = (n – 1, k – 1) + (n – 1, k)

For our example of 3 people selected from a group of 5, we have (5, 3) = (4, 2) + (4, 3)

We’ve broken a large problem into two smaller ones. Instead of selecting from a group of 5, we’re selecting twice from a group of 4: First, all the ways to select 2

people from a group of 4, then all the ways to select 3 people from a group of 4.

There are 6 ways to select 2 people from a group of 4. In the (4, 2) term—which we’ll call the left term—these 6 combinations are

BC, BD, BE, CD, CE, DE

A is the missing group member, so to make three-person teams we precede these combinations with A:

ABC, ABD, ABE, ACD, ACE, ADE

There are four ways to select 3 people from a group of 4. In the (4, 3) term—the right term—we have

BCD, BCE, BDE, CDE

When these 4 combinations from the right term are added to the 6 from the left term, we get the 10 combinations for (5, 3).

You can apply the same decomposition process to each of the groups of 4. For example, (4, 2) is (3, 1) added to (3, 2). As you can see, this is a natural place to apply recursion.

You can think of this problem as a tree with (5,3) on the top row, (4,3) and (4,2) on the next row, and so on, where the nodes in the tree correspond to recursive function calls. Figure 6.20 shows what this looks like for the (5,3) example.

The base cases are combinations that make no sense: those with a 0 for either number and those where the team size is greater than the group size. The combination (1,1) is valid but there’s no point trying to break it down further. In the figure, dotted lines show the base cases; you return rather than following them.

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5, 3

A

4, 2

4, 3

B

B

3, 1

3, 2

3, 2

3, 3

ABC

C

C

2, 0

2, 1

2, 1

2, 2

2, 1

2, 2

2, 2

2, 3

A

AC

B

B

C

D

D

D

D

D

1, –1

1, 0

1, 0

1, 1

1, 0

1, 1

1, 1

1, 2

1, 0

1, 1

1, 1

1, 2

1, 1

1, 2

1, 2

1, 3

ABE

AC

ADE

BCE

BDE

CDE

E

**FIGURE 6.20**

Picking a team of 3 from a group of 5.

The recursion depth corresponds to the group members: The node on the top row represents group member A, the two nodes on the next row represent group member B, and so on. If there are 5 group members, you’ll have 5 levels.

As you descend the tree you need to remember the sequence of members you visit.

Here’s how to do that: Whenever you make a call to a left term, you record the node you’re leaving by adding its letter to a sequence. These left calls and the letters to add to the sequence are shown by the darker lines in the figure. You’ll need to role the sequence back up as you return.

To record all the combinations, you can display them as you go along. You don’t display anything when making left calls. However, when you make calls to the right, you check the sequence; if you’re at a valid node, and adding one member will complete the team, then add the node to the sequence and display the complete team.

**Summary**

• A recursive method calls itself repeatedly, with different argument values each time.

• Some value of its arguments causes a recursive method to return without calling itself. This is called the base case.

• When the innermost instance of a recursive method returns, the process

“unwinds” by completing pending instances of the method, going from the latest back to the original call.

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• A triangular number is the sum of itself and all numbers smaller than itself.

(

is 10, because 4+3+2+1 = 10.

• The factorial of a number is the product of itself and all numbers smaller than itself. For example, the factorial of 4 is 4*3*2*1 = 24.

• Both triangular numbers and factorials can be calculated using either a recursive method or a simple loop.

• The anagram of a word (all possible combinations of its n letters) can be found recursively by repeatedly rotating all its letters and anagramming the rightmost n-1 of them.

• A binary search can be carried out recursively by checking which half of a sorted range the search key is in, and then doing the same thing with that half.

• The Towers of Hanoi puzzle consists of three towers and an arbitrary number of rings.

• The Towers of Hanoi puzzle can be solved recursively by moving all but the bottom disk of a subtree to an intermediate tower, moving the bottom disk to the destination tower, and finally moving the subtree to the destination.

• Merging two sorted arrays means to create a third array that contains all the elements from both arrays in sorted order.

• In mergesort, 1-element subarrays of a larger array are merged into 2-element subarrays, 2-element subarrays are merged into 4-element subarrays, and so on until the entire array is sorted.

• mergesort requires O(N*logN) time.

• mergesort requires a workspace equal in size to the original array.

• For triangular numbers, factorials, anagrams, and the binary search, the recursive method contains only one call to itself. (There are two shown in the code for the binary search, but only one is used on any given pass through the method’s code.)

• For the Towers of Hanoi and mergesort, the recursive method contains two calls to itself.

• Any operation that can be carried out with recursion can be carried out with a stack.

• A recursive approach may be inefficient. If so, it can sometimes be replaced with a simple loop or a stack-based approach.

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**Questions**

These questions are intended as a self-test for readers. Answers may be found in Appendix C.

**1. **If the user enters 10 in the triangle.java program (Listing 6.1), what is the maximum number of “copies” of the triangle() method (actually just copies of its argument) that exist at any one time?

**2. **Where are the copies of the argument, mentioned in question 1, stored?

**a. **in a variable in the triangle() method

**b. **in a field of the TriangleApp class

**c. **in a variable of the getString() method

**d. **on a stack

**3. **Assume the user enters 10 as in question 1. What is the value of n when the triangle() method first returns a value other than 1?

**4. **Assume the same situation as in question 1. What is the value of n when the triangle() method is about to return to main()?

**5. **True or false: In the triangle() method, the return values are stored on the stack.

**6. **In the anagram.java program (Listing 6.2), at a certain depth of recursion, a version of the doAnagram() method is working with the string “led”. When this method calls a new version of itself, what letters will the new version be working with?

**7. **We’ve seen that recursion can take the place of a loop, as in the loop-oriented orderedArray.java program (Listing 2.4) and the recursive binarySearch.java program (Listing 6.3). Which of the following is

**a. **Both programs divide the range repeatedly in half.

**b. **If the key is not found, the loop version returns because the range bounds cross, but the recursive version occurs because it reaches the bottom recursion level.

**c. **If the key is found, the loop version returns from the entire method, whereas the recursive version returns from only one level of recursion.

**d. **In the recursive version the range to be searched must be specified in the arguments, while in the loop version it need not be.

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**8. **In the recFind() method in the binarySearch.java program (Listing 6.3), what takes the place of the loop in the non-recursive version?

**a. **the recFind() method

**b. **arguments to recFind()

**c. **recursive calls to recFind()

**d. **the call from main() to recFind()

**9. **The binarySearch.java program is an example of the _________ approach to solving a problem.

**10. **What gets smaller as you make repeated recursive calls in the redFind() method?

**11. **What becomes smaller with repeated recursive calls in the towers.java program (Listing 6.4)?

**12. **The algorithm in the towers.java program involves **a. **“trees” that are data storage devices.

**b. **secretly putting small disks under large disks.

**c. **changing which columns are the source and destination.

**d. **moving one small disk and then a stack of larger disks.

**13. **Which is

**a. **Its algorithm can handle arrays of different sizes.

**b. **It must search the target array to find where to put the next item.

**c. **It is not recursive.

**d. **It continuously takes the smallest item irrespective of what array it’s in.

**14. **The disadvantage of mergesort is that

**a. **it is not recursive.

**b. **it uses more memory.

**c. **although faster than the insertion sort, it is much slower than quicksort.

**d. **it is complicated to implement.

**15. **Besides a loop, a ___________ can often be used instead of recursion.

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**Experiments**

**1. **In the triangle.java program (Listing 6.1), remove the code for the base case (the if(n==1), the return 1;, and the else). Then run the program and see what happens.

**2. **Use the Towers Workshop applet in manual mode to solve the puzzle with seven or more disks.

**3. **Rewrite the main() part of mergeSort.java (Listing 6.6) so you can fill the array with hundreds of thousands of random numbers. Run the program to sort these numbers and compare its speed with the sorts in Chapter 3, “Simple Sorting.”

**Programming Projects**

(As noted in the Introduction, qualified instructors may obtain completed solutions to the Programming Projects on the publisher’s Web site.) **6.1 **Suppose you buy a budget-priced pocket PC and discover that the chip inside can’t do multiplication, only addition. You program your way out of this quandary by writing a recursive method, mult(), that performs multiplication of x and y by adding x to itself y times. Its arguments are x and y and its return value is the product of x and y. Write such a method and a main() program to call it. Does the addition take place when the method calls itself or when it returns?

**6.2 **In Chapter 8, “Binary Trees,” we’ll look at binary trees, where every branch has (potentially) exactly two sub-branches. If we draw a binary tree on the screen using characters, we might have 1 branch on the top row, 2 on the next row, then 4, 8, 16, and so on. Here’s what that looks like for a tree 16 characters wide:

--------X-------

----X-------X---

--X---X---X---X-

-X-X-X-X-X-X-X-X

XXXXXXXXXXXXXXXX

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(Note that the bottom line should be shifted a half character-width right, but there’s nothing we can do about that with character-mode graphics.) You can draw this tree using a recursive makeBranches() method with arguments left and right, which are the endpoints of a horizontal range. When you first enter the routine, left is 0 and right is the number of characters (including dashes) in all the lines, minus 1. You draw an X in the center of this range. Then the method calls itself twice: once for the left half of the range and once for the right half. Return when the range gets too small. You will probably want to put all the dashes and Xs into an array and display the array all at once, perhaps with a display() method. Write a main() program to draw the tree by calling makeBranches() and display(). Allow main() to determine the line length of the display (32, 64, or whatever). Ensure that the array that holds the characters for display is no larger than it needs to be. What is the relationship of the number of lines (five in the picture here) to the line width?

**6.3 **Implement the recursive approach to raising a number to a power, as described in the “Raising a Number to a Power” section near the end of this chapter.

Write the recursive power() function and a main() routine to test it.

**6.4 **Write a program that solves the knapsack problem for an arbitrary knapsack capacity and series of weights. Assume the weights are stored in an array. Hint: The arguments to the recursive knapsack() function are the target weight and the array index where the remaining items start.

**6.5 **Implement a recursive approach to showing all the teams that can be created from a group (n things taken k at a time). Write the recursive showTeams() method and a main() method to prompt the user for the group size and the team size to provide arguments for showTeam(), which then displays all the possible combinations.

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**7**

**IN THIS CHAPTER**

• Shellsort

Advanced Sorting

• Partitioning

• Quicksort

• Radix Sort

We discussed simple sorting in the aptly titled Chapter 3, “Simple Sorting.” The sorts described there—the bubble, selection, and insertion sorts—are easy to implement but are rather slow. In Chapter 6, “Recursion,” we described the mergesort. It runs much faster than the simple sorts but requires twice as much space as the original array; this is often a serious drawback.

This chapter covers two advanced approaches to sorting: Shellsort and quicksort. These sorts both operate much faster than the simple sorts: the Shellsort in about O(N*(logN)2) time, and quicksort in O(N*logN) time.

Neither of these sorts requires a large amount of extra space, as mergesort does. The Shellsort is almost as easy to implement as mergesort, while quicksort is the fastest of all the general-purpose sorts. We’ll conclude the chapter with a brief mention of the radix sort, an unusual and interesting approach to sorting.

We’ll examine the Shellsort first. Quicksort is based on the idea of partitioning, so we’ll then examine partitioning separately, before examining quicksort itself.

**Shellsort**

The Shellsort is named for Donald L. Shell, the computer scientist who discovered it in 1959. It’s based on the insertion sort, but adds a new feature that dramatically improves the insertion sort’s performance.

The Shellsort is good for medium-sized arrays, perhaps up to a few thousand items, depending on the particular implementation. It’s not quite as fast as quicksort and other O(N*logN) sorts, so it’s not optimum for very large files. However, it’s much faster than the O(N2) sorts like the selection sort and the insertion sort, and it’s very easy to implement: The code is short and simple.

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The worst-case performance is not significantly worse than the average performance.

(We’ll see later in this chapter that the worst-case performance for quicksort can be much worse unless precautions are taken.) Some experts (see Sedgewick in Appendix B, “Further Reading”) recommend starting with a Shellsort for almost any sorting project and changing to a more advanced sort, like quicksort, only if Shellsort proves too slow in practice.

**Insertion Sort: Too Many Copies**

Because Shellsort is based on the insertion sort, you might want to review the section titled “Insertion Sort” in Chapter 3. Recall that partway through the insertion sort the items to the left of a marker are internally sorted (sorted among themselves) and items to the right are not. The algorithm removes the item at the marker and stores it in a temporary variable. Then, beginning with the item to the left of the newly vacated cell, it shifts the sorted items right one cell at a time, until the item in the temporary variable can be reinserted in sorted order.

Here’s the problem with the insertion sort. Suppose a small item is on the far right, where the large items should be. To move this small item to its proper place on the left, all the intervening items (between the place where it is and where it should be) must be shifted one space right. This step takes close to N copies, just for one item.

Not all the items must be moved a full N spaces, but the average item must be moved N/2 spaces, which takes N times N/2 shifts for a total of N2/2 copies. Thus, the performance of insertion sort is O(N2).

This performance could be improved if we could somehow move a smaller item many spaces to the left without shifting all the intermediate items individually.

**N-Sorting**

The Shellsort achieves these large shifts by insertion-sorting widely spaced elements.

After they are sorted, it sorts somewhat less widely spaced elements, and so on. The spacing between elements for these sorts is called the

After 0, 4, and 8 are sorted, the algorithm shifts over one cell and sorts 1, 5, and 9.

This process continues until all the elements have been

After the complete 4-sort, the array can be thought of as comprising four subarrays: (0,4,8), (1,5,9), (2,6), and (3,7), each of which is completely sorted. These subarrays are interleaved but otherwise independent.

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0

1 2 3 4 5 6 7 8 9

Unsorted

0 1 2 3 4 5 6 7 8 9

Sorted

**FIGURE 7.1**

4-sorting 0, 4, and 8.

Notice that, in this particular example, at the end of the 4-sort no item is more than two cells from where it would be if the array were completely sorted. This is what is meant by an array being “almost” sorted and is the secret of the Shellsort. By creating interleaved, internally sorted sets of items, we minimize the amount of work that must be done to complete the sort.

Now, as we noted in Chapter 3, the insertion sort is very efficient when operating on an array that’s almost sorted. If it needs to move items only one or two cells to sort the file, it can operate in almost O(N) time. Thus, after the array has been 4-sorted, we can 1-sort it using the ordinary insertion sort. The combination of the 4-sort and the 1-sort is much faster than simply applying the ordinary insertion sort without the preliminary 4-sort.

**Diminishing Gaps**

We’ve shown an initial interval—or gap—of 4 cells for sorting a 10-cell array. For larger arrays the interval should start out much larger. The interval is then repeatedly reduced until it becomes 1.

For instance, an array of 1,000 items might be 364-sorted, then 121-sorted, then 40-sorted, then 13-sorted, then 4-sorted, and finally 1-sorted. The sequence of numbers used to generate the intervals (in this example, 364, 121, 40, 13, 4, 1) is called the

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attributed to Knuth (see Appendix B), is a popular one. In reversed form, starting from 1, it’s generated by the recursive expression

h = 3*h + 1

where the initial value of h is 1. The first two columns of Table 7.1 show how this formula generates the sequence.

7

10

1

9

2

5

8

6

4

3

0

1

2

3

4

5

6

7

8

9

2

10

1

9

4

5

8

6

7

3

0

1

2

3

4

5

6

7

8

9

2

3

1

9

4

5

8

6

7

10

0

1

2

3

4

5

6

7

8

9

2

3

1

9

4

5

8

6

7

10

0

1

2

3

4

5

6

7

8

9

2

3

1

6

4

5

8

9

7

10

0

1

2

3

4

5

6

7

8

9

**FIGURE 7.2**

A complete 4-sort.

**TABLE 7.1**

Knuth’s Interval Sequence

**h**

**3*h + 1**

**(h–1) / 3**

1

4

4

13

1

13

40

4

40

121

13

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**TABLE 7.1**

Continued

**h**

**3*h + 1**

**(h–1) / 3**

121

364

40

364

1093

121

1093

3280

364

There are other approaches to generating the interval sequence; we’ll return to this issue later. First, we’ll explore how the Shellsort works using Knuth’s sequence.

In the sorting algorithm, the sequence-generating formula is first used in a short loop to figure out the initial gap. A value of 1 is used for the first value of h, and the h=h*3+1 formula is applied to generate the sequence 1, 4, 13, 40, 121, 364, and so on. This process ends when the gap is larger than the array. For a 1,000-element array, the seventh number in the sequence, 1,093, is too large. Thus, we begin the sorting process with the sixth-largest number, creating a 364-sort. Then, each time through the outer loop of the sorting routine, we reduce the interval using the inverse of the formula previously given:

h = (h–1) / 3

This is shown in the third column of Table 7.1. This inverse formula generates the reverse sequence 364, 121, 40, 13, 4, 1. Starting with 364, each of these numbers is used to n-sort the array. When the array has been 1-sorted, the algorithm is done.

**The Shellsort Workshop Applet**

You can use the Shellsort Workshop applet to see how this sort works. Figure 7.3

shows the applet after all the bars have been 4-sorted, just as the 1-sort begins.

**FIGURE 7.3**

The Shellsort Workshop applet.

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As you single-step through the algorithm, you’ll notice that the explanation we gave in the preceding discussion is slightly simplified. The sequence for the 4-sort is not actually (0,4,8), (1,5,9), (2,6), and (3,7). Instead, the first two elements of each group of three are sorted first, then the first two elements of the second group, and so on.

Once the first two elements of all the groups are sorted, the algorithm returns and sorts three-element groups. The actual sequence is (0,4), (1,5), (2,6), (3,7), (0,4,8), (1,5,9).

It might seem more obvious for the algorithm to 4-sort each complete subarray first—(0,4), (0,4,8), (1,5), (1,5,9), (2,6), (3,7)—but the algorithm handles the array indices more efficiently using the first scheme.

The Shellsort is actually not very efficient with only 10 items, making almost as many swaps and comparisons as the insertion sort. However, with 100 bars the improvement becomes significant.

It’s instructive to run the Workshop applet starting with 100 inversely sorted bars.

(Remember that, as in Chapter 3, the first press of New creates a random sequence of bars, while the second press creates an inversely sorted sequence.) Figure 7.4 shows how the bars look after the first pass, when the array has been completely 40-sorted.

Figure 7.5 shows the situation after the next pass, when it is 13-sorted. With each new value of h, the array becomes more nearly sorted.

**FIGURE 7.4**

After the 40-sort.

Why is the Shellsort so much faster than the insertion sort, on which it’s based?

When h is large, the number of items per pass is small, and items move long distances. This is very efficient. As h grows smaller, the number of items per pass increases, but the items are already closer to their final sorted positions, which is

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more efficient for the insertion sort. It’s the combination of these trends that makes the Shellsort so effective.

**FIGURE 7.5**

After the 13-sort.

Notice that later sorts (small values of h) don’t undo the work of earlier sorts (large values of h). An array that has been 40-sorted remains 40-sorted after a 13-sort, for example. If this wasn’t so, the Shellsort couldn’t work.

**Java Code for the Shellsort**

The Java code for the Shellsort is scarcely more complicated than for the insertion sort. Starting with the insertion sort, you substitute h for 1 in appropriate places and add the formula to generate the interval sequence. We’ve made shellSort() a method in the ArraySh class, a version of the array classes from Chapter 2, “Arrays.”

Listing 7.1 shows the complete shellSort.java program.

**LISTING 7.1**

The shellSort.java Program

// shellSort.java

// demonstrates shell sort

// to run this program: C>java ShellSortApp

//--------------------------------------------------------------

class ArraySh

{

private long[] theArray; // ref to array theArray private int nElems; // number of data items

//--------------------------------------------------------------

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**LISTING 7.1**

Continued

public ArraySh(int max) // constructor

{

theArray = new long[max]; // create the array

nElems = 0; // no items yet

}

//--------------------------------------------------------------

public void insert(long value) // put element into array

{

theArray[nElems] = value; // insert it

nElems++; // increment size

}

//--------------------------------------------------------------

public void display() // displays array contents

{

System.out.print(“A=”);

for(int j=0; j<nElems; j++) // for each element, System.out.print(theArray[j] + “ “); // display it System.out.println(“”);

}

//--------------------------------------------------------------

public void shellSort()

{

int inner, outer;

long temp;

int h = 1; // find initial value of h while(h <= nElems/3)

h = h*3 + 1; // (1, 4, 13, 40, 121, ...) while(h>0) // decreasing h, until h=1

{

// h-sort the file

for(outer=h; outer<nElems; outer++)

{

temp = theArray[outer];

inner = outer;

// one subpass (eg 0, 4, 8)

while(inner > h-1 && theArray[inner-h] >= temp)

{

theArray[inner] = theArray[inner-h];

inner -= h;

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**LISTING 7.1**

Continued

}

theArray[inner] = temp;

} // end for

h = (h-1) / 3; // decrease h

} // end while(h>0)

} // end shellSort()

//--------------------------------------------------------------

} // end class ArraySh

////////////////////////////////////////////////////////////////

class ShellSortApp

{

public static void main(String[] args)

{

int maxSize = 10; // array size

ArraySh arr;

arr = new ArraySh(maxSize); // create the array

for(int j=0; j<maxSize; j++) // fill array with

{ // random numbers

long n = (int)(java.lang.Math.random()*99);

arr.insert(n);

}

arr.display(); // display unsorted array arr.shellSort(); // shell sort the array arr.display(); // display sorted array

} // end main()

} // end class ShellSortApp

In main() we create an object of type ArraySh, able to hold 10 items, fill it with random data, display it, Shellsort it, and display it again. Here’s some sample output: A=20 89 6 42 55 59 41 69 75 66

A=6 20 41 42 55 59 66 69 75 89

You can change maxSize to higher numbers, but don’t go too high; 10,000 items take a fraction of a minute to sort.

The Shellsort algorithm, although it’s implemented in just a few lines, is not simple to follow. To see the details of its operation, step through a 10-item sort with the Workshop applet, comparing the messages generated by the applet with the code in the shellSort() method.

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**Other Interval Sequences**

Picking an interval sequence is a bit of a black art. Our discussion so far used the formula h=h*3+1 to generate the interval sequence, but other interval sequences have been used with varying degrees of success. The only absolute requirement is that the diminishing sequence ends with 1, so the last pass is a normal insertion sort.

In Shell’s original paper, he suggested an initial gap of N/2, which was simply divided in half for each pass. Thus, the descending sequence for N=100 is 50, 25, 12, 6, 3, 1. This approach has the advantage that you don’t need to calculate the sequence before the sort begins to find the initial gap; you just divide N by 2.

However, this turns out not to be the best sequence. Although it’s still better than the insertion sort for most data, it sometimes degenerates to O(N2) running time, which is no better than the insertion sort.

A variation of this approach is to divide each interval by 2.2 instead of 2. For n=100

this leads to 45, 20, 9, 4, 1. This is considerably better than dividing by 2, as it avoids some worst-case circumstances that lead to O(N2) behavior. Some extra code is needed to ensure that the last value in the sequence is 1, no matter what N is. This gives results comparable to Knuth’s sequence shown in the listing.

Another possibility for a descending sequence (from Flamig; see Appendix B) is if(h < 5)

h = 1;

else

h = (5*h-1) / 11;

It’s generally considered important that the numbers in the interval sequence are relatively prime; that is, they have no common divisors except 1. This constraint makes it more likely that each pass will intermingle all the items sorted on the previous pass. The inefficiency of Shell’s original N/2 sequence is due to its failure to adhere to this rule.

You may be able to invent a gap sequence of your own that does just as well (or possibly even better) than those shown. Whatever it is, it should be quick to calculate so as not to slow down the algorithm.

**Efficiency of the Shellsort**

No one so far has been able to analyze the Shellsort’s efficiency theoretically, except in special cases. Based on experiments, there are various estimates, which range from 0(N3/2) down to O(N7/6).

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Table 7.2 shows some of these estimated O() values, compared with the slower insertion sort and the faster quicksort. The theoretical times corresponding to various values of N are shown. Note that Nx/y means the yth root of N raised to the x power.

Thus, if N is 100, N3/2 is the square root of 1003, which is 1,000. Also, (logN)2 means the log of N, squared. This is often written log2N, but that’s easy to confuse with log N, the logarithm to the base 2 of N.

2

**TABLE 7.2**

Estimates of Shellsort Running Time

**10**

**100**

**1,000**

**10,000**

**O() Value**

**Type of Sort**

**Items**

**Items**

**Items**

**Items**

N2

Insertion, etc.

100

10,000

1,000,000

100,000,000

N3/2

Shellsort 32

1,000

32,000

1,000,000

N*(logN)2

Shellsort

10

400

9,000

160,000

N5/4

Shellsort 18

316

5,600

100,000

N7/6

Shellsort

14

215

3,200

46,000

N*logN

Quicksort, etc.

10

200

3,000

40,000

For most data, the higher estimates, such as N3/2, are probably more realistic.

**Partitioning**

Partitioning is the underlying mechanism of quicksort, which we’ll explore next, but it’s also a useful operation on its own, so we’ll cover it here in its own section.

To

You can easily imagine situations in which you would want to partition data. Maybe you want to divide your personnel records into two groups: employees who live within 15 miles of the office and those who live farther away. Or a school adminis-trator might want to divide students into those with grade point averages higher and lower than 3.5, so as to know who deserves to be on the Dean’s list.

**The Partition Workshop Applet**

Our Partition Workshop applet demonstrates the partitioning process. Figure 7.6

shows 12 bars before partitioning, and Figure 7.7 shows them again after partitioning.

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**FIGURE 7.6**

Twelve bars before partitioning.

**FIGURE 7.7**

Twelve bars after partitioning.

The horizontal line represents the

This value is returned from the partitioning method, so it can be used by other methods that need to know where the division is.

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For a more vivid display of the partitioning process, set the Partition Workshop applet to 100 bars and press the Run button. The leftScan and rightScan pointers will zip toward each other, swapping bars as they go. When they meet, the partition is complete.

You can choose any value you want for the pivot value, depending on why you’re doing the partition (such as choosing a grade point average of 3.5). For variety, the Workshop applet chooses a random number for the pivot value (the horizontal black line) each time New or Size is pressed, but the value is never too far from the average bar height.

After being partitioned, the data is by no means sorted; it has simply been divided into two groups. However, it’s more sorted than it was before. As we’ll see in the next section, it doesn’t take much more trouble to sort it completely.

Notice that partitioning is not

**The **partition.java **Program**

How is the partitioning process carried out? Let’s look at some example code. Listing 7.2 shows the partition.java program, which includes the partitionIt() method for partitioning an array.

**LISTING 7.2**

The partition.java Program

// partition.java

// demonstrates partitioning an array

// to run this program: C>java PartitionApp

////////////////////////////////////////////////////////////////

class ArrayPar

{

private long[] theArray; // ref to array theArray private int nElems; // number of data items

//--------------------------------------------------------------

public ArrayPar(int max) // constructor

{

theArray = new long[max]; // create the array

nElems = 0; // no items yet

}

//--------------------------------------------------------------

public void insert(long value) // put element into array

{

theArray[nElems] = value; // insert it

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**LISTING 7.2**

Continued

nElems++; // increment size

}

//--------------------------------------------------------------

public int size() // return number of items

{ return nElems; }

//--------------------------------------------------------------

public void display() // displays array contents

{

System.out.print(“A=”);

for(int j=0; j<nElems; j++) // for each element, System.out.print(theArray[j] + “ “); // display it System.out.println(“”);

}

//--------------------------------------------------------------

public int partitionIt(int left, int right, long pivot)

{

int leftPtr = left - 1; // right of first elem int rightPtr = right + 1; // left of pivot

while(true)

{

while(leftPtr < right && // find bigger item theArray[++leftPtr] < pivot)

; // (nop)

while(rightPtr > left && // find smaller item theArray[--rightPtr] > pivot)

; // (nop)

if(leftPtr >= rightPtr) // if pointers cross, break; // partition done

else // not crossed, so

swap(leftPtr, rightPtr); // swap elements

} // end while(true)

return leftPtr; // return partition

} // end partitionIt()

//--------------------------------------------------------------

public void swap(int dex1, int dex2) // swap two elements

{

long temp;

temp = theArray[dex1]; // A into temp

theArray[dex1] = theArray[dex2]; // B into A

theArray[dex2] = temp; // temp into B

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**LISTING 7.2**

Continued

} // end swap()

//--------------------------------------------------------------

} // end class ArrayPar

////////////////////////////////////////////////////////////////

class PartitionApp

{

public static void main(String[] args)

{

int maxSize = 16; // array size

ArrayPar arr; // reference to array arr = new ArrayPar(maxSize); // create the array

for(int j=0; j<maxSize; j++) // fill array with

{ // random numbers

long n = (int)(java.lang.Math.random()*199);

arr.insert(n);

}

arr.display(); // display unsorted array long pivot = 99; // pivot value

System.out.print(“Pivot is “ + pivot);

int size = arr.size();

// partition array

int partDex = arr.partitionIt(0, size-1, pivot);

System.out.println(“, Partition is at index “ + partDex); arr.display(); // display partitioned array

} // end main()

The main() routine creates an ArrayPar object that holds 16 items of type long. The pivot value is fixed at 99. The routine inserts 16 random values into ArrayPar, displays them, partitions them by calling the partitionIt() method, and displays them again. Here’s some sample output:

A=149 192 47 152 159 195 61 66 17 167 118 64 27 80 30 105

Pivot is 99, partition is at index 8

A=30 80 47 27 64 17 61 66 195 167 118 159 152 192 149 105

You can see that the partition is successful: The first eight numbers are all smaller than the pivot value of 99; the last eight are all larger.

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Notice that the partitioning process doesn’t necessarily divide the array in half as it does in this example; that depends on the pivot value and key values of the data.

There may be many more items in one group than in the other.

**The Partition Algorithm**

The partitioning algorithm works by starting with two pointers, one at each end of the array. (We use the term

Actually, leftPtr is initialized to one position to the left of the first cell, and rightPtr to one position to the right of the last cell, because they will be incremented and decremented, respectively, before they’re used.

**Stopping and Swapping**

When leftPtr encounters a data item smaller than the pivot value, it keeps going because that item is already on the correct side of the array. However, when it encounters an item larger than the pivot value, it stops. Similarly, when rightPtr encounters an item larger than the pivot, it keeps going, but when it finds a smaller item, it also stops. Two inner while loops, the first for leftPtr and the second for rightPtr, control the scanning process. A pointer stops because its while loop exits.

Here’s a simplified version of the code that scans for out-of-place items: while( theArray[++leftPtr] < pivot ) // find bigger item

; // (nop)

while( theArray[--rightPtr] > pivot ) // find smaller item

; // (nop)

swap(leftPtr, rightPtr); // swap elements The first while loop exits when an item larger than pivot is found; the second loop exits when an item smaller than pivot is found. When both these loops exit, both leftPtr and rightPtr point to items that are in the wrong sides of the array, so these items are swapped.

After the swap, the two pointers continue on, again stopping at items that are in the wrong side of the array and swapping them. All this activity is nested in an outer while loop, as can be seen in the partitionIt() method in Listing 7.2. When the two pointers eventually meet, the partitioning process is complete and this outer while loop exits.

You can watch the pointers in action when you run the Partition Workshop applet with 100 bars. These pointers, represented by blue arrows, start at opposite ends of

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the array and move toward each other, stopping and swapping as they go. The bars between them are unpartitioned; those they’ve already passed over are partitioned.

When they meet, the entire array is partitioned.

**Handling Unusual Data**

If we were sure that there was a data item at the right end of the array that was smaller than the pivot value, and an item at the left end that was larger, the simplified while loops previously shown would work fine. Unfortunately, the algorithm may be called upon to partition data that isn’t so well organized.

If all the data is smaller than the pivot value, for example, the leftPtr variable will go all the way across the array, looking in vain for a larger item, and fall off the right end, creating an

To avoid these problems, extra tests must be placed in the while loops to check for the ends of the array: leftPtr<right in the first loop and rightPtr>left in the second.

You can see these tests in context in Listing 7.2.

In the section on quicksort, we’ll see that a clever pivot-selection process can eliminate these end-of-array tests. Eliminating code from inner loops is always a good idea if you want to make a program run faster.

**Delicate Code**

The code in the while loops is rather delicate. For example, you might be tempted to remove the increment operators from the inner while loops and use them to replace the nop statements. (Nop refers to a statement consisting only of a semicolon, and means

; // (nop)

to this:

while(leftPtr < right && theArray[leftPtr] < pivot)

++leftPtr;

and similarly for the other inner while loop. These changes would make it possible for the initial values of the pointers to be left and right, which is somewhat clearer than left-1 and right+1.

However, these changes result in the pointers being incremented only when the condition is satisfied. The pointers must move in any case, so two extra statements within the outer while loop would be required to bump the pointers. The nop version is the most efficient solution.

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**Equal Keys**

Here’s another subtle change you might be tempted to make in the partitionIt() code. If you run the partitionIt() method on items that are all equal to the pivot value, you will find that every comparison leads to a swap. Swapping items with equal keys seems like a waste of time. The < and > operators that compare pivot with the array elements in the while loops cause the extra swapping. However, suppose you try to fix this by replacing them with <= and >= operators. This indeed prevents the swapping of equal elements, but it also causes leftPtr and rightPtr to end up at the ends of the array when the algorithm has finished. As we’ll see in the section on quicksort, it’s good for the pointers to end up in the middle of the array, and very bad for them to end up at the ends. So if partitionIt() is going to be used for quicksort, the < and > operators are the right way to go, even if they cause some unnecessary swapping.

**Efficiency of the Partition Algorithm**

The partition algorithm runs in O(N) time. It’s easy to see why this is so when running the Partition Workshop applet: The two pointers start at opposite ends of the array and move toward each other at a more or less constant rate, stopping and swapping as they go. When they meet, the partition is complete. If there were twice as many items to partition, the pointers would move at the same rate, but they would have twice as many items to compare and swap, so the process would take twice as long. Thus, the running time is proportional to N.

More specifically, for each partition there will be N+1 or N+2 comparisons. Every item will be encountered and used in a comparison by one or the other of the pointers, leading to N comparisons, but the pointers overshoot each other before they find out they’ve “crossed” or gone beyond each other, so there are one or two extra comparisons before the partition is complete. The number of comparisons is independent of how the data is arranged (except for the uncertainty between one or two extra comparisons at the end of the scan).

The number of swaps, however, does depend on how the data is arranged. If it’s inversely ordered, and the pivot value divides the items in half, then every pair of values must be swapped, which is N/2 swaps. (Remember in the Partition Workshop applet that the pivot value is selected randomly, so that the number of swaps for inversely sorted bars won’t always be exactly N/2.) For random data, there will be fewer than N/2 swaps in a partition, even if the pivot value is such that half the bars are shorter and half are taller. This is because some bars will already be in the right place (short bars on the left, tall bars on the right). If the pivot value is higher (or lower) than most of the bars, there will be even fewer swaps because only those few bars that are higher (or lower) than the pivot will need to be swapped. On average, for random data, about half the maximum number of swaps take place.

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Although there are fewer swaps than comparisons, they are both proportional to N.

Thus, the partitioning process runs in O(N) time. Running the Workshop applet, you can see that for 12 random bars there are about 3 swaps and 14 comparisons, and for 100 random bars there are about 25 swaps and 102 comparisons.

**Quicksort**

Quicksort is undoubtedly the most popular sorting algorithm, and for good reason: In the majority of situations, it’s the fastest, operating in O(N*logN) time. (This is only true for

To understand quicksort, you should be familiar with the partitioning algorithm described in the preceding section. Basically, the quicksort algorithm operates by partitioning an array into two subarrays and then calling itself recursively to quicksort each of these subarrays. However, there are some embellishments we can make to this basic scheme. They have to do with the selection of the pivot and the sorting of small partitions. We’ll examine these refinements after we’ve looked at a simple version of the main algorithm.

It’s difficult to understand

**The Quicksort Algorithm**

The code for a basic recursive quicksort method is fairly simple. Here’s an example: public void recQuickSort(int left, int right)

{

if(right-left <= 0) // if size is 1,

return; // it’s already sorted

else // size is 2 or larger

{

// partition range

int partition = partitionIt(left, right);

recQuickSort(left, partition-1); // sort left side recQuickSort(partition+1, right); // sort right side

}

}

As you can see, there are three basic steps:

**1. **Partition the array or subarray into left (smaller keys) and right (larger keys) groups.

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**2. **Call ourselves to sort the left group.

**3. **Call ourselves again to sort the right group.

After a partition, all the items in the left subarray are smaller than all those on the right. If we then sort the left subarray and sort the right subarray, the entire array will be sorted. How do we sort these subarrays? By calling ourself recursively.

The arguments to the recQuickSort() method determine the left and right ends of the array (or subarray) it’s supposed to sort. The method first checks if this array consists of only one element. If so, the array is by definition already sorted, and the method returns immediately. This is the base case in the recursion process.

If the array has two or more cells, the algorithm calls the partitionIt() method, described in the preceding section, to partition it. This method returns the index number of the

Unpartitioned array

42

89

63

12

94

27

78

3

50

36

Pivot

Partition

Left subarray

Right subarray

3

27

12

36

63

94

89

78

42

50

Left

Already

Right

Sorted

Will be sorted

Will be sorted

by first recursive

by second recursive

call to recQuickSort()

call to recQuickSort()

**FIGURE 7.8**

Recursive calls sort subarrays.

After the array is partitioned, recQuickSort() calls itself recursively, once for the left part of its array, from left to partition-1, and once for the right, from partition+1 to right. Note that the data item at the index partition is not included in either of the recursive calls. Why not? Doesn’t it need to be sorted? The explanation lies in how the pivot value is chosen.

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**Choosing a Pivot Value**

What pivot value should the partitionIt() method use? Here are some relevant ideas:

• The pivot value should be the key value of an actual data item; this item is called the

• You can pick a data item to be the pivot more or less at random. For simplicity, let’s say we always pick the item on the right end of the subarray being partitioned.

• After the partition, if the pivot is inserted at the boundary between the left and right subarrays, it will be in its final sorted position.

This last point may sound unlikely, but remember that, because the pivot’s key value is used to partition the array, following the partition the left subarray holds items smaller than the pivot, and the right subarray holds items larger. The pivot starts out on the right, but if it could somehow be placed between these two subarrays, it would be in the correct place—that is, in its final sorted position. Figure 7.9 shows how this looks with a pivot whose key value is 36.

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Pivot item

Correct place

for pivot

Partitioned

Partitioned

left subarray

right subarray

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42

50

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**FIGURE 7.9**

The pivot and the subarrays.

This figure is somewhat fanciful because you can’t actually take an array apart as we’ve shown. So how do we move the pivot to its proper place?

We could shift all the items in the right subarray to the right one cell to make room for the pivot. However, this is inefficient and unnecessary. Remember that all the

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items in the right subarray, although they are larger than the pivot, are not yet sorted, so they can be moved around, within the right subarray, without affecting anything. Therefore, to simplify inserting the pivot in its proper place, we can simply swap the pivot (36) and the left item in the right subarray, which is 63. This swap places the pivot in its proper position between the left and right groups. The 63

is switched to the right end, but because it remains in the right (larger) group, the partitioning is undisturbed. This situation is shown in Figure 7.10.

Left subarray

Right subarray

Pivot

3

27

12

63

94

89

78

42

50

36

Left subarray

Right subarray

3

27

12

36

94

89

78

42

50

63

Pivot

**FIGURE 7.10**

Swapping the pivot.

When it’s swapped into the partition’s location, the pivot is in its final resting place.

All subsequent activity will take place on one side of it or on the other, but the pivot itself won’t be moved (or indeed even accessed) again.

To incorporate the pivot selection process into our recQuickSort() method, let’s make it an overt statement, and send the pivot value to partitionIt() as an argument.

Here’s how that looks:

public void recQuickSort(int left, int right)

{

if(right-left <= 0) // if size <= 1,

return; // already sorted

else // size is 2 or larger

{

long pivot = theArray[right]; // rightmost item

// partition range

int partition = partitionIt(left, right, pivot);

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recQuickSort(left, partition-1); // sort left side recQuickSort(partition+1, right); // sort right side

}

} // end recQuickSort()

When we use this scheme of choosing the rightmost item in the array as the pivot, we’ll need to modify the partitionIt() method to exclude this rightmost item from the partitioning process; after all, we already know where it should go after the partitioning process is complete: at the partition, between the two groups. Also, after the partitioning process is completed, we need to swap the pivot from the right end into the partition’s location. Listing 7.3 shows the quickSort1.java program, which incorporates these features.

**LISTING 7.3**

The quickSort1.java Program

// quickSort1.java

// demonstrates simple version of quick sort

// to run this program: C>java QuickSort1App

////////////////////////////////////////////////////////////////

class ArrayIns

{

private long[] theArray; // ref to array theArray private int nElems; // number of data items

//--------------------------------------------------------------

public ArrayIns(int max) // constructor

{

theArray = new long[max]; // create the array

nElems = 0; // no items yet

}

//--------------------------------------------------------------

public void insert(long value) // put element into array

{

theArray[nElems] = value; // insert it

nElems++; // increment size

}

//--------------------------------------------------------------

public void display() // displays array contents

{

System.out.print(“A=”);

}

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**LISTING 7.3**

Continued

//--------------------------------------------------------------

public void quickSort()

{

recQuickSort(0, nElems-1);

}

//--------------------------------------------------------------

public void recQuickSort(int left, int right)

{

if(right-left <= 0) // if size <= 1, return; // already sorted

else // size is 2 or larger

{

long pivot = theArray[right]; // rightmost item

// partition range

int partition = partitionIt(left, right, pivot);

recQuickSort(left, partition-1); // sort left side recQuickSort(partition+1, right); // sort right side

}

} // end recQuickSort()

//--------------------------------------------------------------

public int partitionIt(int left, int right, long pivot)

{

int leftPtr = left-1; // left (after ++) int rightPtr = right; // right-1 (after --) while(true)

{ // find bigger item

while( theArray[++leftPtr] < pivot )

; // (nop)

// find smaller item

while(rightPtr > 0 && theArray[--rightPtr] > pivot)

; // (nop)

if(leftPtr >= rightPtr) // if pointers cross, break; // partition done

else // not crossed, so

swap(leftPtr, rightPtr); // swap elements

} // end while(true)

swap(leftPtr, right); // restore pivot

return leftPtr; // return pivot location

} // end partitionIt()

//--------------------------------------------------------------

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**LISTING 7.3**

Continued

public void swap(int dex1, int dex2) // swap two elements

{

long temp = theArray[dex1]; // A into temp

theArray[dex1] = theArray[dex2]; // B into A

theArray[dex2] = temp; // temp into B

} // end swap(

//--------------------------------------------------------------

} // end class ArrayIns

////////////////////////////////////////////////////////////////

class QuickSort1App

{

public static void main(String[] args)

{

int maxSize = 16; // array size

ArrayIns arr;

arr = new ArrayIns(maxSize); // create array

for(int j=0; j<maxSize; j++) // fill array with

{ // random numbers

long n = (int)(java.lang.Math.random()*99);

arr.insert(n);

}

arr.display(); // display items

arr.quickSort(); // quicksort them

arr.display(); // display them again

} // end main()

} // end class QuickSort1App

The main() routine creates an object of type ArrayIns, inserts 16 random data items of type long in it, displays it, sorts it with the quickSort() method, and displays the results. Here’s some typical output:

A=69 0 70 6 38 38 24 56 44 26 73 77 30 45 97 65

A=0 6 24 26 30 38 38 44 45 56 65 69 70 73 77 97

An interesting aspect of the code in the partitionIt() method is that we’ve been able to remove the test for the end of the array in the first inner while loop. This test, seen in the earlier partitionIt() method in the partition.java program in Listing 7.2, was

leftPtr < right

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It prevented leftPtr running off the right end of the array if no item there was larger than pivot. Why can we eliminate the test? Because we selected the rightmost item as the pivot, so leftPtr will always stop there. However, the test is still necessary for rightPtr in the second while loop. (Later we’ll see how this test can be eliminated as well.)

Choosing the rightmost item as the pivot is thus not an entirely arbitrary choice; it speeds up the code by removing an unnecessary test. Picking the pivot from some other location would not provide this advantage.

**The QuickSort1 Workshop Applet**

At this point you know enough about the quicksort algorithm to understand the nuances of the QuickSort1 Workshop applet.

**The Big Picture**

For the big picture, use the Size button to set the applet to sort 100 random bars, and press the Run button. Following the sorting process, the display will look something like Figure 7.11.

**FIGURE 7.11**

The QuickSort1 Workshop applet with 100 bars.

Watch how the algorithm partitions the array into two parts, then sorts each of these parts by partitioning it into two parts, and so on, creating smaller and smaller subarrays.

When the sorting process is complete, each dotted line provides a visual record of one of the sorted subarrays. The horizontal range of the line shows which bars were part of the subarray, and its vertical position is the pivot value (the height of the

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pivot). The total length of all these lines on the display is a measure of how much work the algorithm has done to sort the array; we’ll return to this topic later.

Each dotted line (except the shortest ones) should have a line below it (probably separated by other, shorter lines) and a line above it that together add up to the same length as the original line (less one bar). These are the two partitions into which each subarray is divided.

**The Details**

For a more detailed examination of quicksort’s operation, switch to the 12-bar display in the QuickSort1 Workshop applet and step through the sorting process.

You’ll see how the pivot value corresponds to the height of the pivot on the right side of the array and how the algorithm partitions the array, swaps the pivot into the space between the two sorted groups, sorts the shorter group (using many recursive calls), and then sorts the larger group.

Figure 7.12 shows all the steps involved in sorting 12 bars. The horizontal brackets under the arrays show which subarray is being partitioned at each step, and the circled numbers show the order in which these partitions are created. A pivot being swapped into place is shown with a dotted arrow. The final position of the pivot is shown as a dotted cell to emphasize that this cell contains a sorted item that will not be changed thereafter. Horizontal brackets under single cells (steps 5, 6, 7, 11, and 12) are base case calls to recQuickSort(); they return immediately.

Sometimes, as in steps 4 and 10, the pivot ends up in its original position on the right side of the array being sorted. In this situation, there is only one subarray remaining to be sorted: the one to the left of the pivot. There is no second subarray to its right.

The different steps in Figure 7.12 occur at different levels of recursion, as shown in Table 7.3. The initial call from main() to recQuickSort() is the first level, recQuickSort() calling two new instances of itself is the second level, these two instances calling four more instances is the third level, and so on.

**TABLE 7.3**

Recursion Levels for Figure 7.12

**Step**

**Recursion Level**

1

1

2, 8

2

3, 7, 9, 12

3

4, 10

4

5, 6, 11

5

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Fig 7-2

11

12

**FIGURE 7.12**

The quicksort process.

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The order in which the partitions are created, corresponding to the step numbers, does not correspond with depth. It’s not the case that all the first-level partitions are done first, then all the second level ones, and so on. Instead, the left group at every level is handled before any of the right groups.

In theory there should be 8 steps in the fourth level and 16 in the fifth level, but in this small array we run out of items before these steps are necessary.

The number of levels in the table shows that with 12 data items, the machine stack needs enough space for 5 sets of arguments and return values; one for each recursion level. This is, as we’ll see later, somewhat greater than the logarithm to the base 2 of the number of items: log N. The size of the machine stack is determined by your 2

particular system. Sorting very large numbers of data items using recursive procedures may cause this stack to overflow, leading to memory errors.

**Things to Notice**

Here are some details you may notice as you run the QuickSort1 Workshop applet.

You might think that a powerful algorithm like quicksort would not be able to handle subarrays as small as two or three items. However, this version of the quicksort algorithm is quite capable of sorting such small subarrays; leftScan and rightScan just don’t go very far before they meet. For this reason we don’t need to use a different sorting scheme for small subarrays. (Although, as we’ll see later, handling small subarrays differently may have advantages.) At the end of each scan, the leftScan variable ends up pointing to the partition—

that is, the left element of the right subarray. The pivot is then swapped with the partition to put the pivot in its proper place, as we’ve seen. As we noted, in steps 3

and 9 of Figure 7.12, leftScan ends up pointing to the pivot itself, so the swap has no effect. This may seem like a wasted swap; you might decide that leftScan should stop one bar sooner. However, it’s important that leftScan scan all the way to the pivot; otherwise, a swap would unsort the pivot and the partition.

Be aware that leftScan and rightScan start at left-1 and right. This may look peculiar on the display, especially if left is 0; then leftScan will start at –1. Similarly, rightScan initially points to the pivot, which is not included in the partitioning process. These pointers start outside the subarray being partitioned because they will be incremented and decremented, respectively, before they’re used the first time.

The applet shows ranges as numbers in parentheses; for example, (2-5) means the subarray from index 2 to index 5. The range given in some of the messages may be negative: from a higher number to a lower one, such as

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int partition = partitionIt(left, right, pivot);

recQuickSort(left, partition-1); // sort left side recQuickSort(partition+1, right); // sort right side If partitionIt() is called with left = 7 and right = 8, for example, and happens to return 7 as the partition, then the range supplied in the first call to recQuickSort() will be (7-6) and the range to the second will be (8-8). This is normal. The base case in recQuickSort() is activated by array sizes less than 1 as well as by 1, so it will return immediately for negative ranges. Negative ranges are not shown in Figure 7.12, although they do cause (brief) calls to recQuickSort().

**Degenerates to O(N2) Performance**

If you use the QuickSort1 Workshop applet to sort 100 inversely sorted bars, you’ll see that the algorithm runs much more slowly and that many more dotted horizontal lines are generated, indicating more and larger subarrays are being partitioned.

What’s happening here?

The problem is in the selection of the pivot. Ideally, the pivot should be the median of the items being sorted. That is, half the items should be larger than the pivot, and half smaller. This would result in the array being partitioned into two subarrays of equal size. Having two equal subarrays is the optimum situation for the quicksort algorithm. If it has to sort one large and one small array, it’s less efficient because the larger subarray has to be subdivided more times.

The worst situation results when a subarray with N elements is divided into one subarray with 1 element and the other with N-1 elements. (This division into 1 cell and N-1 cells can also be seen in steps 3 and 9 in Figure 7.12.) If this 1 and N-1 division happens with every partition, then every element requires a separate partition step. This is in fact what takes place with inversely sorted data: In all the subarrays, the pivot is the smallest item, so every partition results in N-1 elements in one subarray and only the pivot in the other.

To see this unfortunate process in action, step through the QuickSort1 Workshop applet with 12 inversely sorted bars. Notice how many more steps are necessary than with random data. In this situation the advantage gained by the partitioning process is lost and the performance of the algorithm degenerates to O(N2).

Besides being slow, there’s another potential problem when quicksort operates in O(N2) time. When the number of partitions increases, the number of recursive function calls also increases. Every function call takes up room on the machine stack. If there are too many calls, the machine stack may overflow and paralyze the system.

To summarize: In the QuickSort1 applet, we select the rightmost element as the pivot. If the data is truly random, this isn’t too bad a choice because usually the

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pivot won’t be too close to either end of the array. However, when the data is sorted or inversely sorted, choosing the pivot from one end or the other is a bad idea. Can we improve on our approach to selecting the pivot?

**Median-of-Three Partitioning**

Many schemes have been devised for picking a better pivot. The method should be simple but have a good chance of avoiding the largest or smallest value. Picking an element at random is simple but—as we’ve seen—doesn’t always result in a good selection. However, we could examine all the elements and actually calculate which one was the median. This would be the ideal pivot choice, but the process isn’t practical, as it would take more time than the sort itself.

A compromise solution is to find the median of the first, last, and middle elements of the array, and use this for the pivot. Picking the median of the first, last, and middle elements is called the

Left

Center

Right

44

86

29

Median is 44

**FIGURE 7.13**

The median of three.

Finding the median of three items is obviously much faster than finding the median of all the items, and yet it successfully avoids picking the largest or smallest item in cases where the data is already sorted or inversely sorted. There are probably some pathological arrangements of data where the median-of-three scheme works poorly, but normally it’s a fast and effective technique for finding the pivot.

Besides picking the pivot more effectively, the median-of-three approach has an additional benefit: We can dispense with the rightPtr>left test in the second inside while loop, leading to a small increase in the algorithm’s speed. How is this possible?

The test can be eliminated because we can use the median-of-three approach to not only select the pivot, but also to sort the three elements used in the selection process. Figure 7.14 shows this operation.

When these three elements are sorted, and the median item is selected as the pivot, we are guaranteed that the element at the left end of the subarray is less than (or equal to) the pivot, and the element at the right end is greater than (or equal to) the

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pivot. This means that the leftPtr and rightPtr indices can’t step beyond the right or left ends of the array, respectively, even if we remove the leftPtr>right and rightPtr<left tests. (The pointer will stop, thinking it needs to swap the item, only to find that it has crossed the other pointer and the partition is complete.) The values at left and right act as

Left

Center

Right

44

86

29

Before sorting

Left

Center

Right

29

44

86

After sorting

Becomes

pivot

**FIGURE 7.14**

Sorting the left, center, and right elements.

Another small benefit to median-of-three partitioning is that after the left, center, and right elements are sorted, the partition process doesn’t need to examine these elements again. The partition can begin at left+1 and right-1 because left and right have in effect already been partitioned. We know that left is in the correct partition because it’s on the left and it’s less than the pivot, and right is in the correct place because it’s on the right and it’s greater than the pivot.

Thus, median-of-three partitioning not only avoids O(N2) performance for already-sorted data, it also allows us to speed up the inner loops of the partitioning algorithm and reduce slightly the number of items that must be partitioned.

**The **quickSort2.java **Program**

Listing 7.4 shows the quickSort2.java program, which incorporates median-of-three partitioning. We use a separate method, medianOf3(), to sort the left, center, and right elements of a subarray. This method returns the value of the pivot, which is then sent to the partitionIt() method.

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**LISTING 7.4**

The quickSort2.java Program

// quickSort2.java

// demonstrates quick sort with median-of-three partitioning

// to run this program: C>java QuickSort2App

////////////////////////////////////////////////////////////////

class ArrayIns

{

private long[] theArray; // ref to array theArray private int nElems; // number of data items

//--------------------------------------------------------------

public ArrayIns(int max) // constructor

{

theArray = new long[max]; // create the array

nElems = 0; // no items yet

}

//--------------------------------------------------------------

public void insert(long value) // put element into array

{

theArray[nElems] = value; // insert it

nElems++; // increment size

}

//--------------------------------------------------------------

public void display() // displays array contents

{

System.out.print(“A=”);

}

//--------------------------------------------------------------

public void quickSort()

{

recQuickSort(0, nElems-1);

}

//--------------------------------------------------------------

public void recQuickSort(int left, int right)

{

int size = right-left+1;

if(size <= 3) // manual sort if small manualSort(left, right);

else // quicksort if large

{

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**LISTING 7.4**

Continued

long median = medianOf3(left, right);

int partition = partitionIt(left, right, median);

recQuickSort(left, partition-1);

recQuickSort(partition+1, right);

}

} // end recQuickSort()

//--------------------------------------------------------------

public long medianOf3(int left, int right)

{

int center = (left+right)/2;

// order left & center

if( theArray[left] > theArray[center] )

swap(left, center);

// order left & right

if( theArray[left] > theArray[right] )

swap(left, right);

// order center & right

if( theArray[center] > theArray[right] )

swap(center, right);

swap(center, right-1); // put pivot on right return theArray[right-1]; // return median value

} // end medianOf3()

//--------------------------------------------------------------

public void swap(int dex1, int dex2) // swap two elements

{

long temp = theArray[dex1]; // A into temp

theArray[dex1] = theArray[dex2]; // B into A

theArray[dex2] = temp; // temp into B

} // end swap(

//--------------------------------------------------------------

public int partitionIt(int left, int right, long pivot)

{

int leftPtr = left; // right of first elem int rightPtr = right - 1; // left of pivot

while(true)

{

while( theArray[++leftPtr] < pivot ) // find bigger

; // (nop)

while( theArray[--rightPtr] > pivot ) // find smaller

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**LISTING 7.4**

Continued

; // (nop)

if(leftPtr >= rightPtr) // if pointers cross, break; // partition done

else // not crossed, so

swap(leftPtr, rightPtr); // swap elements

} // end while(true)

swap(leftPtr, right-1); // restore pivot

return leftPtr; // return pivot location

} // end partitionIt()

//--------------------------------------------------------------

public void manualSort(int left, int right)

{

int size = right-left+1;

if(size <= 1)

return; // no sort necessary

if(size == 2)

{ // 2-sort left and right

if( theArray[left] > theArray[right] )

swap(left, right);

return;

}

else // size is 3

{ // 3-sort left, center, & right if( theArray[left] > theArray[right-1] )

swap(left, right-1); // left, center if( theArray[left] > theArray[right] )

swap(left, right); // left, right

if( theArray[right-1] > theArray[right] )

swap(right-1, right); // center, right

}

} // end manualSort()

//--------------------------------------------------------------

} // end class ArrayIns

////////////////////////////////////////////////////////////////

class QuickSort2App

{

public static void main(String[] args)

{

int maxSize = 16; // array size

ArrayIns arr; // reference to array arr = new ArrayIns(maxSize); // create the array

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**LISTING 7.4**

Continued

for(int j=0; j<maxSize; j++) // fill array with

{ // random numbers

long n = (int)(java.lang.Math.random()*99);

arr.insert(n);

}

arr.display(); // display items

arr.quickSort(); // quicksort them

arr.display(); // display them again

} // end main()

} // end class QuickSort2App

This program uses another new method, manualSort(), to sort subarrays of three or fewer elements. It returns immediately if the subarray is one cell (or less), swaps the cells if necessary if the range is 2, and sorts three cells if the range is 3. The recQuickSort() routine can’t be used to sort ranges of 2 or 3 because median partitioning requires at least four cells.

The main() routine and the output of quickSort2.java are similar to those of quickSort1.java.

**The QuickSort2 Workshop Applet**

The Quicksort2 Workshop applet demonstrates the quicksort algorithm using median-of-three partitioning. This applet is similar to the QuickSort1 Workshop applet, but starts off sorting the first, center, and left elements of each subarray and selecting the median of these as the pivot value. At least, it does this if the array size is greater than 3. If the subarray is two or three units, the applet simply sorts it “by hand” without partitioning or recursive calls.

Notice the dramatic improvement in performance when the applet is used to sort 100 inversely ordered bars. No longer is every subarray partitioned into 1 cell and N-1 cells; instead, the subarrays are partitioned roughly in half.

Other than this improvement for ordered data, the QuickSort2 Workshop applet produces results similar to QuickSort1. It is no faster when sorting random data; it’s advantages become evident only when sorting ordered data.

**Handling Small Partitions**

If you use the median-of-three partitioning method, it follows that the quicksort algorithm won’t work for partitions of three or fewer items. The number 3 in this case is called a

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**Using an Insertion Sort for Small Partitions**

Another option for dealing with small partitions is to use the insertion sort. When you do this, you aren’t restricted to a cutoff of 3. You can set the cutoff to 10, 20, or any other number. It’s interesting to experiment with different values of the cutoff to see where the best performance lies. Knuth (see Appendix B) recommends a cutoff of 9. However, the optimum number depends on your computer, operating system, compiler (or interpreter), and so on.

The quickSort3.java program, shown in Listing 7.5, uses an insertion sort to handle subarrays of fewer than 10 cells.

**LISTING 7.5**

The quickSort3.java Program

// quickSort3.java

// demonstrates quick sort; uses insertion sort for cleanup

// to run this program: C>java QuickSort3App

////////////////////////////////////////////////////////////////

class ArrayIns

{

private long[] theArray; // ref to array theArray private int nElems; // number of data items

//--------------------------------------------------------------

public ArrayIns(int max) // constructor

{

theArray = new long[max]; // create the array

nElems = 0; // no items yet

}

//--------------------------------------------------------------

public void insert(long value) // put element into array

{

theArray[nElems] = value; // insert it

nElems++; // increment size

}

//--------------------------------------------------------------

public void display() // displays array contents

{

System.out.print(“A=”);

}

//--------------------------------------------------------------

public void quickSort()

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**LISTING 7.5**

Continued

{

recQuickSort(0, nElems-1);

// insertionSort(0, nElems-1); // the other option

}

//--------------------------------------------------------------

public void recQuickSort(int left, int right)

{

int size = right-left+1;

if(size < 10) // insertion sort if small insertionSort(left, right);

else // quicksort if large

{

long median = medianOf3(left, right);

int partition = partitionIt(left, right, median);

recQuickSort(left, partition-1);

recQuickSort(partition+1, right);

}

} // end recQuickSort()

//--------------------------------------------------------------

public long medianOf3(int left, int right)

{

int center = (left+right)/2;

// order left & center

if( theArray[left] > theArray[center] )

swap(left, center);

// order left & right

if( theArray[left] > theArray[right] )

swap(left, right);

// order center & right

if( theArray[center] > theArray[right] )

swap(center, right);

swap(center, right-1); // put pivot on right return theArray[right-1]; // return median value

} // end medianOf3()

//--------------------------------------------------------------

public void swap(int dex1, int dex2) // swap two elements

{

long temp = theArray[dex1]; // A into temp

theArray[dex1] = theArray[dex2]; // B into A

theArray[dex2] = temp; // temp into B

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**LISTING 7.5**

Continued

} // end swap(

//--------------------------------------------------------------

public int partitionIt(int left, int right, long pivot)

{

int leftPtr = left; // right of first elem int rightPtr = right - 1; // left of pivot

while(true)

{

while( theArray[++leftPtr] < pivot ) // find bigger

; // (nop)

while( theArray[--rightPtr] > pivot ) // find smaller

; // (nop)

if(leftPtr >= rightPtr) // if pointers cross, break; // partition done

else // not crossed, so

swap(leftPtr, rightPtr); // swap elements

} // end while(true)

swap(leftPtr, right-1); // restore pivot

return leftPtr; // return pivot location

} // end partitionIt()

//--------------------------------------------------------------

// insertion sort

public void insertionSort(int left, int right)

{

int in, out;

// sorted on left of out

for(out=left+1; out<=right; out++)

{

long temp = theArray[out]; // remove marked item in = out; // start shifts at out

// until one is smaller,

while(in>left && theArray[in-1] >= temp)

{

theArray[in] = theArray[in-1]; // shift item to right

--in; // go left one position

}

theArray[in] = temp; // insert marked item

} // end for

} // end insertionSort()

//--------------------------------------------------------------

} // end class ArrayIns

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**LISTING 7.5**

Continued

////////////////////////////////////////////////////////////////

class QuickSort3App

{

public static void main(String[] args)

{

int maxSize = 16; // array size

ArrayIns arr; // reference to array arr = new ArrayIns(maxSize); // create the array

for(int j=0; j<maxSize; j++) // fill array with

{ // random numbers

long n = (int)(java.lang.Math.random()*99);

arr.insert(n);

}

arr.display(); // display items

arr.quickSort(); // quicksort them

arr.display(); // display them again

} // end main()

Using the insertion sort for small subarrays turns out to be the fastest approach on our particular installation, but it is not much faster than sorting subarrays of three or fewer cells by hand, as in quickSort2.java. The numbers of comparisons and copies are reduced substantially in the quicksort phase, but are increased by an almost equal amount in the insertion sort, so the time savings are not dramatic. However, this approach is probably worthwhile if you are trying to squeeze the last ounce of performance out of quicksort.

**Insertion Sort Following Quicksort**

Another option is to completely quicksort the array without bothering to sort partitions smaller than the cutoff. This is shown with a commented-out line in the quickSort() method. (If this call is used, the call to insertionSort() should be removed from recQuickSort().) When quicksort is finished, the array will be almost sorted. You then apply the insertion sort to the entire array. The insertion sort is supposed to operate efficiently on almost-sorted arrays, and this approach is recommended by some experts, but on our installation it runs very slowly. The insertion sort appears to be happier doing a lot of small sorts than one big one.

**Removing Recursion**

Another embellishment recommended by many writers is removing recursion from the quicksort algorithm. This involves rewriting the algorithm to store deferred

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subarray bounds (left and right) on a stack, and using a loop instead of recursion to oversee the partitioning of smaller and smaller subarrays. The idea in doing this is to speed up the program by removing method calls. However, this idea arose with older compilers and computer architectures, which imposed a large time penalty for each method call. It’s not clear that removing recursion is much of an improvement for modern systems, which handle method calls more efficiently.

**Efficiency of Quicksort**

We’ve said that quicksort operates in O(N*logN) time. As we saw in the discussion of mergesort in Chapter 6, this is generally true of the divide-and-conquer algorithms, in which a recursive method divides a range of items into two groups and then calls itself to handle each group. In this situation the logarithm actually has a base of 2: The running time is proportional to N*log N.

2

You can get an idea of the validity of this N*log N running time for quicksort by 2

running one of the quickSort Workshop applets with 100 random bars and examining the resulting dotted horizontal lines.

Each dotted line represents an array or subarray being partitioned: the pointers leftScan and rightScan moving toward each other, comparing each data item and swapping when appropriate. We saw in the “Partitioning” section that a single partition runs in O(N) time. This tells us that the total length of all the dotted lines is proportional to the running time of quicksort. But how long are all the lines?

Measuring them with a ruler on the screen would be tedious, but we can visualize them a different way.

There is always 1 line that runs the entire width of the graph, spanning N bars. This results from the first partition. There will also be 2 lines (one below and one above the first line) that have an average length of N/2 bars; together they are again N bars long. Then there will be 4 lines with an average length of N/4 that again total N

bars, then 8 lines, 16 lines, and so on. Figure 7.15 shows how this looks for 1, 2, 4, and 8 lines.

In this figure solid horizontal lines represent the dotted horizontal lines in the quicksort applets, and captions like

The circled numbers on the left show the order in which the lines are created.

Each series of lines (the eight N/8 lines, for example) corresponds to a level of recursion. The initial call to recQuickSort() is the first level and makes the first line; the two calls from within the first call—the second level of recursion—make the next two lines; and so on. If we assume we start with 100 cells, the results are shown in Table 7.4.

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15

13

14

long

9

12

cells

10

11

N/4 cells long

N/8

N/2 cells long

N cells long

1

8

lines

Two lines

One line

6

Four lines

7

Eight

2

5

3

4

**FIGURE 7.15**

Lines correspond to partitions.

**TABLE 7.4**

Line Lengths and Recursion

**Step**

**Average**

**Numbers**

**Line**

**Recursion**

**in Figure**

**Length**

**Number of**

**Total Length**

**Level**

**7.15**

**(Cells)**

**Lines**

**(Cells)**

1

1

100

1

100

2

2, 9

50

2

100

3

3, 6, 10,

25

4

100

13

4

4, 5, 7,

12

8

96

8, 11, 12,

14, 15

5

Not shown

6

16

96

6

Not shown

3

32

96

7

Not shown

1

64

64

Total = 652

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Where does this division process stop? If we keep dividing 100 by 2, and count how many times we do this, we get the series 100, 50, 25, 12, 6, 3, 1, which is about seven levels of recursion. This looks about right on the workshop applets: If you pick some point on the graph and count all the dotted lines directly above and below it, there will be an average of approximately seven. (In Figure 7.15, because not all levels of recursion are shown, only four lines intersect any vertical slice of the graph.)

Table 7.4 shows a total of 652 cells. This is only an approximation because of round-off errors, but it’s close to 100 times the logarithm to the base 2 of 100, which is 6.65. Thus, this informal analysis suggests the validity of the N*log N running time 2

for quicksort.

More specifically, in the section on partitioning, we found that there should be N+2

comparisons and fewer than N/2 swaps. Multiplying these quantities by log N for 2

various values of N gives the results shown in Table 7.5.

**TABLE 7.5**

Swaps and Comparisons in Quicksort

**N**

**8**

**12**

**16**

**64**

**100**

**128**

log N

3

3.59

4

6

6.65

7

2

N*log N

24

43

64

384

665

896

2

Comparisons: (N+2)*log N

30

50

72

396

678

910

2

Swaps: fewer than N/2*log N

12

21

32

192

332

448

2

The log N quantity used in Table 7.5 is actually true only in the best-case scenario, 2

where each subarray is partitioned exactly in half. For random data the figure is slightly greater. Nevertheless, the QuickSort1 and QuickSort2 Workshop applets approximate these results for 12 and 100 bars, as you can see by running them and observing the Swaps and Comparisons fields.

Because they have different cutoff points and handle the resulting small partitions differently, QuickSort1 performs fewer swaps but more comparisons than QuickSort2.

The number of swaps shown in Table 7.5 is the maximum (which assumes the data is inversely sorted). For random data the actual number of swaps turns out to be one-half to two-thirds of the figures shown.

**Radix Sort**

We’ll close this chapter by briefly mentioning a sort that uses a different approach.

The sorts we’ve looked at so far treat the key as a simple numerical value that is compared with other values to sort the data. The radix sort disassembles the key into digits and arranges the data items according to the value of the digits. Amazingly, no comparisons are necessary.

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**Algorithm for the Radix Sort**

We’ll discuss the radix sort in terms of normal base-10 arithmetic, which is easier to visualize. However, an efficient implementation of the radix sort would use base-2

arithmetic to take advantage of the computer’s speed in bit manipulation. We’ll look at the

**1. **All the data items are divided into 10 groups, according to the value of their 1s digit.

**2. **These 10 groups are then reassembled: All the keys ending with 0 go first, followed by all the keys ending in 1, and so on up to 9. We’ll call these steps a sub-sort.

**3. **In the second sub-sort, all data is divided into 10 groups again, but this time according to the value of their 10s digit. This must be done without changing the order of the previous sort. That is, within each of the 10 groups, the ordering of the items remains the same as it was after step 2; the sub-sorts must be stable.

**4. **Again the 10 groups are recombined, those with a 10s digit of 0 first, then those with a 10s digit of 1, and so on up to 9.

**5. **This process is repeated for the remaining digits. If some keys have fewer digits than others, their higher-order digits are considered to be 0.

Here’s an example, using seven data items, each with three digits. Leading zeros are shown for clarity.

421 240 035 532 305 430 124 // unsorted array (240 430) (421) (532) (124) (035 305) // sorted on 1s digit (305) (421 124) (430 532 035) (240) // sorted on 10s digit (035) (124) (240) (305) (421 430) (532) // sorted on 100s digit 035 124 240 305 421 430 532 // sorted array The parentheses delineate the groups. Within each group the digits in the appropriate position are the same. To convince yourself that this approach really works, try it on a piece of paper with some numbers you make up.

**Designing a Program**

In practice the original data probably starts out in an ordinary array. Where should the 10 groups go? There’s a problem with using another array or an array of 10

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arrays. It’s not likely there will be exactly the same number of 0s, 1s, 2s, and so on in every digit position, so it’s hard to know how big to make the arrays. One way to solve this problem is to use 10 linked lists instead of 10 arrays. Linked lists expand and contract as needed. We’ll use this approach.

An outer loop looks at each digit of the keys in turn. There are two inner loops: The first takes the data from the array and puts it on the lists; the second copies it from the lists back to the array. You need to use the right kind of linked list. To keep the sub-sorts stable, you need the data to come out of each list in the same order it went in. Which kind of linked list makes this easy? We’ll leave the coding details as an exercise.

**Efficiency of the Radix Sort**

At first glance the efficiency of the radix sort seems too good to be true. All you do is copy the original data from the array to the lists and back again. If there are 10 data items, this is 20 copies. You repeat this procedure once for each digit. If you assume, say, 5-digit numbers, then you’ll have 20*5 equals 100 copies. If you have 100 data items, there are 200*5 equals 1,000 copies. The number of copies is proportional to the number of data items, which is O(N), the most efficient sorting algorithm we’ve seen.

Unfortunately, it’s generally true that if you have more data items, you’ll need longer keys. If you have 10 times as much data, you may need to add another digit to the key. The number of copies is proportional to the number of data items times the number of digits in the key. The number of digits is the log of the key values, so in most situations we’re back to O(N*logN) efficiency, the same as quicksort.

There are no comparisons, although it takes time to extract each digit from the number. This must be done once for every two copies. It may be, however, that a given computer can do the digit-extraction in binary more quickly than it can do a comparison. Of course, like mergesort, the radix sort uses about twice as much memory as quicksort.

**Summary**

• The Shellsort applies the insertion sort to widely spaced elements, then less widely spaced elements, and so on.

• The expression

• A sequence of numbers, called the

• A widely used interval sequence is generated by the recursive expression h=3*h+1, where the initial value of h is 1.

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• If an array holds 1,000 items, it could be 364-sorted, 121-sorted, 40-sorted, 13-sorted, 4-sorted, and finally 1-sorted.

• The Shellsort is hard to analyze, but runs in approximately O(N*(logN)2) time.

This is much faster than the O(N2) algorithms like insertion sort, but slower than the O(N*logN) algorithms like quicksort.

• To

• The

• In the partitioning algorithm, two array indices, each in its own while loop, start at opposite ends of the array and step toward each other, looking for items that need to be swapped.

• When an index finds an item that needs to be swapped, its while loop exits.

• When both while loops exit, the items are swapped.

• When both while loops exit, and the indices have met or passed each other, the partition is complete.

• Partitioning operates in linear O(N) time, making N plus 1 or 2 comparisons and fewer than N/2 swaps.

• The partitioning algorithm may require extra tests in its inner while loops to prevent the indices running off the ends of the array.

• Quicksort partitions an array and then calls itself twice recursively to sort the two resulting subarrays.

• Subarrays of one element are already sorted; this can be a base case for quicksort.

• The pivot value for a partition in quicksort is the key value of a specific item, called the pivot.

• In a simple version of quicksort, the pivot can always be the item at the right end of the subarray.

• During the partition the pivot is placed out of the way on the right, and is not involved in the partitioning process.

• Later the pivot is swapped again, into the space between the two partitions.

This is its final sorted position.

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• In the simple version of quicksort, performance is only O(N2) for already-sorted (or inversely sorted) data.

• In a more advanced version of quicksort, the pivot can be the median of the first, last, and center items in the subarray. This is called

• Median-of-three partitioning effectively eliminates the problem of O(N2) performance for already-sorted data.

• In median-of-three partitioning, the left, center, and right items are sorted at the same time the median is determined.

• This sort eliminates the need for the end-of-array tests in the inner while loops in the partitioning algorithm.

• Quicksort operates in O(N*log N) time (except when the simpler version is 2

applied to already-sorted data).

• Subarrays smaller than a certain size (the

• The insertion sort is commonly used to sort subarrays smaller than the cutoff.

• The insertion sort can also be applied to the entire array, after it has been sorted down to a cutoff point by quicksort.

• The radix sort is about as fast as quicksort but uses twice as much memory.

**Questions**

These questions are intended as a self-test for readers. Answers may be found in Appendix C.

**1. **The Shellsort works by

**a. **partitioning the array.

**b. **swapping adjacent elements.

**c. **dealing with widely separated elements.

**d. **starting with the normal insertion sort.

**2. **If an array has 100 elements, then Knuth’s algorithm would start with an interval of ________.

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**3. **To transform the insertion sort into the Shellsort, which of the following do you

**a. **Substitute h for 1.

**b. **Insert an algorithm for creating gaps of decreasing width.

**c. **Enclose the normal insertion sort in a loop.

**d. **Change the direction of the indices in the inner loop.

**4. **True or false: A good interval sequence for the Shellsort is created by repeatedly dividing the array size in half.

**5. **Fill in the big O values: The speed of the Shellsort is more than _______ but less than ________.

**6. **Partitioning is

**a. **putting all elements larger than a certain value on one end of the array.

**b. **dividing an array in half.

**c. **partially sorting parts of an array.

**d. **sorting each half of an array separately.

**7. **When partitioning, each array element is compared to the _______.

**8. **In partitioning, if an array element is equal to the answer to question 7, **a. **it is passed over.

**b. **it is passed over or not, depending on the other array element.

**c. **it is placed in the pivot position.

**d. **it is swapped.

**9. **True or false: In quicksort, the pivot can be an arbitrary element of the array.

**10. **Assuming larger keys on the right, the partition is **a. **the element between the left and right subarrays.

**b. **the key value of the element between the left and right subarrays.

**c. **the left element in the right subarray.

**d. **the key value of the left element in the right subarray.

**11. **Quicksort involves partitioning the original array and then _________.

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**12. **After a partition in a simple version of quicksort, the pivot may be **a. **used to find the median of the array.

**b. **exchanged with an element of the right subarray.

**c. **used as the starting point of the next partition.

**d. **discarded.

**13. **Median-of-three partitioning is a way of choosing the _______ .

**14. **In quicksort, for an array of N elements, the partitionIt() method will examine each element approximately ______ times.

**15. **True or false: You can speed up quicksort if you stop partitioning when the partition size is 5 and finish by using a different sort.

**Experiments**

**1. **Find out what happens when you use the Partition Workshop applet on 100

inversely sorted bars. Is the result almost sorted?

**2. **Modify the shellSort.java program (Listing 7.1) so it prints the entire contents of the array after completing each n-sort. The array should be small enough so its contents fit on one line. Analyze these intermediate steps to see if the algorithm is operating the way you think should.

**3. **Modify the shellSort.java (Listing 7.1) and the quickSort3.java (Listing 7.5) programs to sort appropriately large arrays, and compare their speeds. Also, compare these speeds with those of the sorts in Chapter 3.

**Programming Projects**

(As noted in the Introduction, qualified instructors may obtain completed solutions to the Programming Projects on the publisher’s Web site.) **7.1 **Modify the partition.java program (Listing 7.2) so that the partitionIt() method always uses the highest-index (right) element as the pivot, rather than an arbitrary number. (This is similar to what happens in the quickSort1.java program in Listing 7.3.) Make sure your routine will work for arrays of three or fewer elements. To do so, you may need a few extra statements.

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**7.2 **Modify the quickSort2.java program (Listing 7.4) to count the number of copies and comparisons it makes during a sort and then display the totals. This program should duplicate the performance of the QuickSort2 Workshop applet, so the copies and comparisons for inversely sorted data should agree.

(Remember that a swap is three copies.)

**7.3 **In Exercise 3.2 in Chapter 3, we suggested that you could find the median of a set of data by sorting the data and picking the middle element. You might think using quicksort and picking the middle element would be the fastest way to find the median, but there’s an even faster way. It uses the partition algorithm to find the median without completely sorting the data.

To see how this works, imagine that you partition the data, and, by chance, the pivot happens to end up at the middle element. You’re done! All the items to the right of the pivot are larger (or equal), and all the items to the left are smaller (or equal), so if the pivot falls in the exact center of the array, then it’s the median. The pivot won’t end up in the center very often, but we can fix that by repartitioning the partition that contains the middle element.

Suppose your array has seven elements numbered from 0 to 6. The middle is element 3. If you partition this array and the pivot ends up at 4, then you need to partition again from 0 to 4 (the partition that contains 3), not 5 to 6. If the pivot ends up at 2, you need to partition from 2 to 6, not 0 to 1. You continue partitioning the appropriate partitions recursively, always checking if the pivot falls on the middle element. Eventually, it will, and you’re done. Because you need fewer partitions than in quicksort, this algorithm is faster.

Extend Programming Project 7.1 to find the median of an array. You’ll make recursive calls somewhat like those in quicksort, but they will only partition each subarray, not completely sort it. The process stops when the median is found, not when the array is sorted.

**7.4 **Selection means finding the kth largest or kth smallest element from an array.

For example, you might want to select the 7th largest element. Finding the median (as in Programming Project 7.2) is a special case of selection. The same partitioning process can be used, but you look for an element with a specified index number rather than the middle element. Modify the program from Programming Project 7.2 to allow the selection of an arbitrary element. How small an array can your program handle?

**7.5 **Implement a radix sort as described in the last section of this chapter. It should handle variable amounts of data and variable numbers of digits in the key. You could make the number-base variable as well (so it can be something other than 10), but it will be hard to see what’s happening unless you develop a routine to print values in different bases.

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**8**

**IN THIS CHAPTER**

• Why Use Binary Trees?

Binary Trees

• Tree Terminology

• An Analogy

• How Do Binary Search Trees

In this chapter we switch from algorithms, the focus of Work?

Chapter 7, “Advanced Sorting,” to data structures. Binary trees are one of the fundamental data storage structures

• Finding a Node

used in programming. They provide advantages that the data structures we’ve seen so far cannot. In this chapter

• Inserting a Node

we’ll learn why you would want to use trees, how they

• Traversing the Tree

work, and how to go about creating them.

• Finding Maximum and

Minimum Values

**Why Use Binary Trees? **

• Deleting a Node

Why might you want to use a tree? Usually, because it combines the advantages of two other structures: an

• The Efficiency of Binary Trees

ordered array and a linked list. You can search a tree

•

quickly, as you can an ordered array, and you can also Trees Represented as Arrays

insert and delete items quickly, as you can with a linked

• Duplicate Keys

list. Let’s explore these topics a bit before delving into the details of trees.

• The Complete tree.java

Program

**Slow Insertion in an Ordered Array**

• The Huffman Code

Imagine an array in which all the elements are arranged in order—that is, an ordered array, such as we saw in Chapter 2, “Arrays.” As we learned, you can quickly search such an array for a particular value, using a binary search. You check in the center of the array; if the object you’re looking for is greater than what you find there, you narrow your search to the top half of the array; if it’s less, you narrow your search to the bottom half. Applying this process repeatedly finds the object in O(logN) time. You can also quickly iterate through an ordered array, visiting each object in sorted order.

On the other hand, if you want to insert a new object into an ordered array, you first need to find where the object will go, and then move all the objects with greater keys up

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one space in the array to make room for it. These multiple moves are time-consuming, requiring, on the average, moving half the items (N/2 moves). Deletion involves the same multimove operation and is thus equally slow.

If you’re going to be doing a lot of insertions and deletions, an ordered array is a bad choice.

**Slow Searching in a Linked List**

On the other hand, as we saw in Chapter 5, “Linked Lists,” insertions and deletions are quick to perform on a linked list. They are accomplished simply by changing a few references. These operations require O(1) time (the fastest Big O time).

Unfortunately, however,

(Notice that times considered fast for a sort are slow for data structure operations.) You might think you could speed things up by using an ordered linked list, in which the elements were arranged in order, but this doesn’t help. You still must start at the beginning and visit the elements in order, because there’s no way to access a given element without following the chain of references to it. (Of course, in an ordered list it’s much quicker to visit the nodes in order than it is in a non-ordered list, but that doesn’t help to find an arbitrary object.)

**Trees to the Rescue**

It would be nice if there were a data structure with the quick insertion and deletion of a linked list, and also the quick searching of an ordered array. Trees provide both these characteristics, and are also one of the most interesting data structures.

**What Is a Tree? **

We’ll be mostly interested in a particular kind of tree called a

A tree consists of

Trees have been studied extensively as abstract mathematical entities, so there’s a large amount of theoretical knowledge about them. A tree is actually an instance of a more general category called a

We’ll discuss graphs in Chapter 13, “Graphs,” and Chapter 14, “Weighted Graphs.”

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367

Nodes

Edges

**FIGURE 8.1**

A general (non-binary) tree.

In computer programs, nodes often represent such entities as people, car parts, airline reservations, and so on—in other words, the typical items we store in any kind of data structure. In an OOP language like Java these real-world entities are represented by objects.

The lines (edges) between the nodes represent the way the nodes are related.

Roughly speaking, the lines represent convenience: It’s easy (and fast) for a program to get from one node to another if there is a line connecting them. In fact, the

Edges are likely to be represented in a program by references, if the program is written in Java (or by pointers if the program is written in C or C++).

Typically, there is one node in the top row of a tree, with lines connecting to more nodes on the second row, even more on the third, and so on. Thus, trees are small on the top and large on the bottom. This may seem upside-down compared with real trees, but generally a program starts an operation at the small end of the tree, and it’s (arguably) more natural to think about going from top to bottom, as in reading text.

There are different kinds of trees. The tree shown in Figure 8.1 has more than two children per node. (We’ll see what “children” means in a moment.) However, in this chapter we’ll be discussing a specialized form of tree called a

**Tree Terminology**

Many terms are used to describe particular aspects of trees. You need to know a few of them so our discussion will be comprehensible. Fortunately, most of these terms are related to real-world trees or to family relationships (as in parents and children),

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so they’re not hard to remember. Figure 8.2 shows many of these terms applied to a binary tree.

Root

A

The dashed

Level 0

line is a path

B is the

parent of D

B

and E

C

Level 1

D is the

left child

E is the

of B

right child

of B

D

E

F

G

Level 2

A subtree

H

I

J

Level 3

with F as

its root

H, E, I, J, and G are leaf nodes

**FIGURE 8.2**

Tree terms.

**Path**

Think of someone walking from node to node along the edges that connect them.

The resulting sequence of nodes is called a

**Root**

The node at the top of the tree is called the

**FIGURE 8.3**

A non-tree.

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369

**Parent**

Any node (except the root) has exactly one edge running upward to another node.

The node above it is called the

**Child**

Any node may have one or more lines running downward to other nodes. These nodes below a given node are called its

**Leaf**

A node that has no children is called a

**Subtree**

Any node may be considered to be the root of a

**Visiting**

A node is

**Traversing**

To

**Levels**

The

**Keys**

We’ve seen that one data field in an object is usually designated a

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**Binary Trees**

If every node in a tree can have at most two children, the tree is called a

In this chapter we’ll focus on binary trees because they are the simplest and the most common.

The two children of each node in a binary tree are called the

The kind of binary tree we’ll be dealing with in this discussion is technically called a

53

30

72

14

39

61

84

9

23

34

47

79

**FIGURE 8.4**

A binary search tree.

**NOTE**

The defining characteristic of a binary search tree is this: A node’s left child must have a key less than its parent, and a node’s right child must have a key greater than or equal to its parent.

**An Analogy**

One commonly encountered tree is the hierarchical file structure in a computer system. The root directory of a given device (designated with the backslash, as in C:\, on many systems) is the tree’s root. The directories one level below the root directory are its children. There may be many levels of subdirectories. Files represent leaves; they have no children of their own.

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371

Clearly, a hierarchical file structure is not a binary tree, because a directory may have many children. A complete pathname, such as C:\SALES\EAST\NOVEMBER\SMITH.DAT, corresponds to the path from the root to the SMITH.DAT leaf. Terms used for the file structure, such as

A hierarchical file structure differs in a significant way from the trees we’ll be discussing here. In the file structure, subdirectories contain no data; they contain only references to other subdirectories or to files. Only files contain data. In a tree, every node contains data (a personnel record, car-part specifications, or whatever). In addition to the data, all nodes except leaves contain references to other nodes.

**How Do Binary Search Trees Work? **

Let’s see how to carry out the common binary tree operations of finding a node with a given key, inserting a new node, traversing the tree, and deleting a node. For each of these operations we’ll first show how to use the Binary Tree Workshop applet to carry it out; then we’ll look at the corresponding Java code.

**The Binary Tree Workshop Applet**

Start up the Binary Tree Workshop applet. You’ll see a screen something like that shown in Figure 8.5. However, because the tree in the Workshop applet is randomly generated, it won’t look exactly the same as the tree in the figure.

**FIGURE 8.5**

The Binary Tree Workshop applet.

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**Using the Workshop Applet**

The key values shown in the nodes range from 0 to 99. Of course, in a real tree, there would probably be a larger range of key values. For example, if employees’

Social Security numbers were used for key values, they would range up to 999,999,999.

Another difference between the Workshop applet and a real tree is that the Workshop applet is limited to a depth of five; that is, there can be no more than five levels from the root to the bottom. This restriction ensures that all the nodes in the tree will be visible on the screen. In a real tree the number of levels is unlimited (until you run out of memory).

Using the Workshop applet, you can create a new tree whenever you want. To do this, click the Fill button. A prompt will ask you to enter the number of nodes in the tree. This can vary from 1 to 31, but 15 will give you a representative tree. After typing in the number, press Fill twice more to generate the new tree. You can experiment by creating trees with different numbers of nodes.

**Unbalanced Trees**

Notice that some of the trees you generate are

90

42

95

Unbalanced

23

75

subtree

10

31

83

7

18

78

87

**FIGURE 8.6**

An unbalanced tree (with an unbalanced subtree).

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373

Trees become unbalanced because of the order in which the data items are inserted.

If these key values are inserted randomly, the tree will be more or less balanced.

However, if an ascending sequence (like 11, 18, 33, 42, 65, and so on) or a descending sequence is generated, all the values will be right children (if ascending) or left children (if descending) and the tree will be unbalanced. The key values in the Workshop applet are generated randomly, but of course some short ascending or descending sequences will be created anyway, which will lead to local imbalances.

When you learn how to insert items into the tree in the Workshop applet, you can try building up a tree by inserting such an ordered sequence of items and see what happens.

If you ask for a large number of nodes when you use Fill to create a tree, you may not get as many nodes as you requested. Depending on how unbalanced the tree becomes, some branches may not be able to hold a full number of nodes. This is because the depth of the applet’s tree is limited to five; the problem would not arise in a real tree.

If a tree is created by data items whose key values arrive in random order, the problem of unbalanced trees may not be too much of a problem for larger trees because the chances of a long run of numbers in sequence is small. But key values can arrive in strict sequence; for example, when a data-entry person arranges a stack of personnel files into order of ascending employee number before entering the data.

When this happens, tree efficiency can be seriously degraded. We’ll discuss unbalanced trees and what to do about them in Chapter 9, “Red-Black Trees.”

**Representing the Tree in Java Code**

Let’s see how we might implement a binary tree in Java. As with other data structures, there are several approaches to representing a tree in the computer’s memory.

The most common is to store the nodes at unrelated locations in memory, and connect them using references in each node that point to its children.

You can also represent a tree in memory as an array, with nodes in specific positions stored in corresponding positions in the array. We’ll return to this possibility at the end of this chapter. For our sample Java code we’ll use the approach of connecting the nodes using references.

**NOTE**

As we discuss individual operations, we’ll show code fragments pertaining to that operation.

The complete program from which these fragments are extracted can be seen toward the end of this chapter in Listing 8.1.

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**The **Node **Class**

First, we need a class of node objects. These objects contain the data representing the objects being stored (employees in an employee database, for example) and also references to each of the node’s two children. Here’s how that looks: class Node

{

int iData; // data used as key value double fData; // other data

node leftChild; // this node’s left child node rightChild; // this node’s right child public void displayNode()

{

// (see Listing 8.1 for method body)

}

}

Some programmers also include a reference to the node’s parent. This simplifies some operations but complicates others, so we don’t include it. We do include a method called displayNode() to display the node’s data, but its code isn’t relevant here.

There are other approaches to designing class Node. Instead of placing the data items directly into the node, you could use a reference to an object representing the data item:

class Node

{

person p1; // reference to person object node leftChild; // this node’s left child node rightChild; // this node’s right child

}

class person

{

int iData;

double fData;

}

This approach makes it conceptually clearer that the node and the data item it holds aren’t the same thing, but it results in somewhat more complicated code, so we’ll stick to the first approach.

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**The **Tree **Class**

We’ll also need a class from which to instantiate the tree itself: the object that holds all the nodes. We’ll call this class Tree. It has only one field: a Node variable that holds the root. It doesn’t need fields for the other nodes because they are all accessed from the root.

The Tree class has a number of methods. They are used for finding, inserting, and deleting nodes; for different kinds of traverses; and for displaying the tree. Here’s a skeleton version:

class Tree

{

private Node root; // the only data field in Tree public void find(int key)

{

}

public void insert(int id, double dd)

{

}

public void delete(int id)

{

}

// various other methods

} // end class Tree

**The **TreeApp **Class**

Finally, we need a way to perform operations on the tree. Here’s how you might write a class with a main() routine to create a tree, insert three nodes into it, and then search for one of them. We’ll call this class TreeApp: class TreeApp

{

public static void main(String[] args)

{

Tree theTree = new Tree; // make a tree

theTree.insert(50, 1.5); // insert 3 nodes

theTree.insert(25, 1.7);

theTree.insert(75, 1.9);

node found = theTree.find(25); // find node with key 25

if(found != null)

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System.out.println(“Found the node with key 25”);

else

System.out.println(“Could not find node with key 25”);

} // end main()

} // end class TreeApp

**TIP**

In Listing 8.1 the main() routine also provides a primitive user interface so you can decide from the keyboard whether you want to insert, find, delete, or perform other operations.

Next we’ll look at individual tree operations: finding a node, inserting a node, traversing the tree, and deleting a node.

**Finding a Node**

Finding a node with a specific key is the simplest of the major tree operations, so let’s start with that.

Remember that the nodes in a binary search tree correspond to objects containing information. They could be

**Using the Workshop Applet to Find a Node**

Look at the Workshop applet, and pick a node, preferably one near the bottom of the tree (as far from the root as possible). The number shown in this node is its

For purposes of this discussion we’ll assume you’ve decided to find the node representing the item with key value 57, as shown in Figure 8.7. Of course, when you run the Workshop applet, you’ll get a different tree and will need to pick a different key value.

Click the Find button. The prompt will ask for the value of the node to find. Enter 57

(or whatever the number is on the node you chose). Click Find twice more.

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57 < 63

63

27

80

57 > 27

57 > 51

13

51

70

92

57 < 58

26

33

58

82

57 == 57

57

60

**FIGURE 8.7**

Finding node 57.

As the Workshop applet looks for the specified node, the prompt will display either Going to left child or Going to right child, and the red arrow will move down one level to the right or left.

In Figure 8.7 the arrow starts at the root. The program compares the key value 57

with the value at the root, which is 63. The key is less, so the program knows the desired node must be on the left side of the tree—either the root’s left child or one of this child’s descendants. The left child of the root has the value 27, so the comparison of 57 and 27 will show that the desired node is in the right subtree of 27. The arrow will go to 51, the root of this subtree. Here, 57 is again greater than the 51

node, so we go to the right, to 58, and then to the left, to 57. This time the comparison shows 57 equals the node’s key value, so we’ve found the node we want.

The Workshop applet doesn’t do anything with the node after finding it, except to display a message saying it has been found. A serious program would perform some operation on the found node, such as displaying its contents or changing one of its fields.

**Java Code for Finding a Node**

Here’s the code for the find() routine, which is a method of the Tree class: public Node find(int key) // find node with given key

{ // (assumes non-empty tree) Node current = root; // start at root

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while(current.iData != key) // while no match,

{

if(key < current.iData) // go left?

current = current.leftChild;

else

current = current.rightChild; // or go right?

if(current == null) // if no child,

return null; // didn’t find it

}

return current; // found it

}

This routine uses a variable current to hold the node it is currently examining. The argument key is the value to be found. The routine starts at the root. (It has to; this is the only node it can access directly.) That is, it sets current to the root.

Then, in the while loop, it compares the value to be found, key, with the value of the iData field (the key field) in the current node. If key is less than this field, current is set to the node’s left child. If key is greater than (or equal) to the node’s iData field, current is set to the node’s right child.

**Can’t Find the Node**

If current becomes equal to null, we couldn’t find the next child node in the sequence; we’ve reached the end of the line without finding the node we were looking for, so it can’t exist. We return null to indicate this fact.

**Found the Node**

If the condition of the while loop is not satisfied, so that we exit from the bottom of the loop, the iData field of current is equal to key; that is, we’ve found the node we want. We return the node so that the routine that called find() can access any of the node’s data.

**Tree Efficiency**

As you can see, the time required to find a node depends on how many levels down it is situated. In the Workshop applet there can be up to 31 nodes, but no more than five levels—so you can find any node using a maximum of only five comparisons.

This is O(logN) time, or more specifically O(log N) time, the logarithm to the base 2.

2

We’ll discuss this further toward the end of this chapter.

**Inserting a Node**

To insert a node, we must first find the place to insert it. This is much the same process as trying to find a node that turns out not to exist, as described in the “Can’t

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379

Find the Node” section. We follow the path from the root to the appropriate node, which will be the parent of the new node. When this parent is found, the new node is connected as its left or right child, depending on whether the new node’s key is less or greater than that of the parent.

**Using the Workshop Applet to Insert a Node**

To insert a new node with the Workshop applet, press the Ins button. You’ll be asked to type the key value of the node to be inserted. Let’s assume we’re going to insert a new node with the value 45. Type this number into the text field.

The first step for the program in inserting a node is to find where it should be inserted. Figure 8.8a shows how this step looks.

60

60

40

40

30

50

30

50

null

45

a) Before insertion

b) After insertion

**FIGURE 8.8**

Inserting a node.

The value 45 is less than 60 but greater than 40, so we arrive at node 50. Now we want to go left because 45 is less than 50, but 50 has no left child; its leftChild field is null. When it sees this null, the insertion routine has found the place to attach the new node. The Workshop applet does this by creating a new node with the value 45

(and a randomly generated color) and connecting it as the left child of 50, as shown in Figure 8.8b.

**Java Code for Inserting a Node**

The insert() function starts by creating the new node, using the data supplied as arguments.

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Next, insert() must determine where to insert the new node. This is done using roughly the same code as finding a node, described in the section “Java Code for Finding a Node.” The difference is that when you’re simply trying to

The value to be searched for is the data item passed in the argument id. The while loop uses true as its condition because it doesn’t care if it encounters a node with the same value as id; it treats another node with the same key value as if it were simply greater than the key value. (We’ll return to the subject of duplicate nodes later in this chapter.)

A place to insert a new node will always be found (unless you run out of memory); when it is, and the new node is attached, the while loop exits with a return statement.

Here’s the code for the insert() function:

public void insert(int id, double dd)

{

Node newNode = new Node(); // make new node

newNode.iData = id; // insert data

newNode.dData = dd;

if(root==null) // no node in root

root = newNode;

else // root occupied

{

Node current = root; // start at root

Node parent;

while(true) // (exits internally)

{

parent = current;

if(id < current.iData) // go left?

{

current = current.leftChild;

if(current == null) // if end of the line,

{ // insert on left

parent.leftChild = newNode;

return;

}

} // end if go left

else // or go right?

{

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current = current.rightChild;

if(current == null) // if end of the line

{ // insert on right

parent.rightChild = newNode;

return;

}

} // end else go right

} // end while

} // end else not root

} // end insert()

// -------------------------------------------------------------

We use a new variable, parent (the parent of current), to remember the last non-null node we encountered (50 in Figure 8.8). This is necessary because current is set to null in the process of discovering that its previous value did not have an appropriate child. If we didn’t save parent, we would lose track of where we were.

To insert the new node, change the appropriate child pointer in parent (the last non-null node you encountered) to point to the new node. If you were looking unsuccessfully for parent’s left child, you attach the new node as parent’s left child; if you were looking for its right child, you attach the new node as its right child. In Figure 8.8, 45 is attached as the left child of 50.

**Traversing the Tree**

Traversing a tree means visiting each node in a specified order. This process is not as commonly used as finding, inserting, and deleting nodes. One reason for this is that traversal is not particularly fast. But traversing a tree is useful in some circumstances, and it’s theoretically interesting. (It’s also simpler than deletion, the discussion of which we want to defer as long as possible.)

There are three simple ways to traverse a tree. They’re called

**Inorder Traversal**

An inorder traversal of a binary search tree will cause all the nodes to be visited in

The simplest way to carry out a traversal is the use of recursion (discussed in Chapter 6, “Recursion”). A recursive method to traverse the entire tree is called with a node as an argument. Initially, this node is the root. The method needs to do only three things:

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**1. **Call itself to traverse the node’s left subtree.

**2. **Visit the node.

**3. **Call itself to traverse the node’s right subtree.

Remember that

Traversals work with any binary tree, not just with binary search trees. The traversal mechanism doesn’t pay any attention to the key values of the nodes; it only concerns itself with whether a node has children.

**Java Code for Traversing**

The actual code for inorder traversal is so simple we show it before seeing how traversal looks in the Workshop applet. The routine, inOrder(), performs the three steps already described. The visit to the node consists of displaying the contents of the node. Like any recursive function, it must have a base case—the condition that causes the routine to return immediately, without calling itself. In inOrder() this happens when the node passed as an argument is null. Here’s the code for the inOrder() method:

private void inOrder(node localRoot)

{

if(localRoot != null)

{

inOrder(localRoot.leftChild);

System.out.print(localRoot.iData + “ “);

inOrder(localRoot.rightChild);

}

}

This method is initially called with the root as an argument: inOrder(root);

After that, it’s on its own, calling itself recursively until there are no more nodes to visit.

**Traversing a Three-Node Tree**

Let’s look at a simple example to get an idea of how this recursive traversal routine works. Imagine traversing a tree with only three nodes: a root (A), with a left child (B), and a right child (C), as shown in Figure 8.9.

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A

B

C

inOrder (A)

1. Call inOrder (B)

2. Visit A

3. Call inOrder (C)

inOrder (B)

inOrder (C)

1. Call inOrder (null)

1. Call inOrder (null)

2. Visit B

2. Visit C

3. Call inOrder (null)

3. Call inOrder (null)

inOrder (null)

inOrder (null)

inOrder (null)

inOrder (null)

Returns

Returns

Returns

Returns

**FIGURE 8.9**

The inOrder() method applied to a three-node tree.

We start by calling inOrder() with the root A as an argument. This incarnation of inOrder() we’ll call inOrder(A). inOrder(A) first calls inOrder() with its left child, B, as an argument. This second incarnation of inOrder() we’ll call inOrder(B).

inOrder(B) now calls itself with its left child as an argument. However, it has no left child, so this argument is null. This creates an invocation of inorder() we could call inOrder(null). There are now three instances of inOrder() in existence: inOrder(A), inOrder(B), and inOrder(null). However, inOrder(null) returns immediately when it finds its argument is null. (We all have days like that.) Now inOrder(B) goes on to visit B; we’ll assume this means to display it. Then inOrder(B) calls inOrder() again, with its right child as an argument. Again this argument is null, so the second inorder(null) returns immediately. Now inOrder(B) has carried out steps 1, 2, and 3, so it returns (and thereby ceases to exist).

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Now we’re back to inOrder(A), just returning from traversing A’s left child. We visit A and then call inOrder() again with C as an argument, creating inOrder(C). Like inOrder(B), inOrder(C) has no children, so step 1 returns with no action, step 2 visits C, and step 3 returns with no action. inOrder(B) now returns to inOrder(A).

However, inOrder(A) is now done, so it returns and the entire traversal is complete.

The order in which the nodes were visited is A, B, C; they have been visited

In a binary search tree this would be the order of ascending keys.

More complex trees are handled similarly. The inOrder() function calls itself for each node, until it has worked its way through the entire tree.

**Traversing with the Workshop Applet**

To see what a traversal looks like with the Workshop applet, repeatedly press the Trav button. (You don’t need to type in any numbers.)

Here’s what happens when you use the Tree Workshop applet to traverse inorder the tree shown in Figure 8.10. This is slightly more complex than the three-node tree seen previously. The red arrow starts at the root. Table 8.1 shows the sequence of node keys and the corresponding messages. The key sequence is displayed at the bottom of the Workshop applet screen.

13. Visit 50

50

1

18

16. Visit 60

7. Visit 30

14

30

60

12

2

15

17

6 8

10. Visit 40

20

40

3

5

9

11

4. Visit 20

**FIGURE 8.10**

Traversing a tree inorder.

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**TABLE 8.1**

Workshop Applet Traversal

**Red**

**List of**

**Step**

**Arrow**

**Nodes**

**Number**

**on Node**

**Message**

**Visited**

1

50 (root)

Will check left child

2

30

Will check left child

3

20

Will check left child

4

20

Will visit this node

5

20

Will check right child

20

6

20

Will go to root of previous subtree

20

7

30

Will visit this node

20

8

30

Will check right child

20 30

9

40

Will check left child

20 30

10

40

Will visit this node

20 30

11

40

Will check right child

20 30 40

12

40

Will go to root of previous subtree

20 30 40

13

50

Will visit this node

20 30 40

14

50

Will check right child

20 30 40 50

15

60

Will check left child

20 30 40 50

16

60

Will visit this node

20 30 40 50

17

60

Will check right child

20 30 40 50 60

18

60

Will go to root of previous subtree

20 30 40 50 60

19

50

Done traversal

20 30 40 50 60

It may not be obvious, but for each node, the routine traverses the node’s left subtree, visits the node, and traverses the right subtree. For example, for node 30 this happens in steps 2, 7, and 8.

The traversal algorithm isn’t as complicated as it looks. The best way to get a feel for what’s happening is to traverse a variety of different trees with the Workshop applet.

**Preorder and Postorder Traversals**

You can traverse the tree in two ways besides inorder; they’re called preorder and postorder. It’s fairly clear why you might want to traverse a tree inorder, but the motivation for preorder and postorder traversals is more obscure. However, these traversals are indeed useful if you’re writing programs that

A binary tree (not a binary search tree) can be used to represent an algebraic expression that involves the binary arithmetic operators +, –, /, and *. The root node holds an operator, and the other nodes hold either a variable name (like A, B, or C), or another operator. Each subtree is a valid algebraic expression.

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*

A

+

B

C

Infix: A*(B+C)

Prefix: *A+BC

Postfix: ABC+*

**FIGURE 8.11**

Tree representing an algebraic expression.

For example, the binary tree shown in Figure 8.11 represents the algebraic expression A*(B+C)

This is called

“Stacks and Queues.”) Traversing the tree inorder will generate the correct inorder sequence A*B+C, but you’ll need to insert the parentheses yourself.

What does all this have to do with preorder and postorder traversals? Let’s see what’s involved. For these other traversals the same three steps are used as for inorder, but in a different sequence. Here’s the sequence for a preorder() method: **1. **Visit the node.

**2. **Call itself to traverse the node’s left subtree.

**3. **Call itself to traverse the node’s right subtree.

Traversing the tree shown in Figure 8.11 using preorder would generate the expression

*A+BC

This is called

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The third kind of traversal, postorder, contains the three steps arranged in yet another way:

**1. **Call itself to traverse the node’s left subtree.

**2. **Call itself to traverse the node’s right subtree.

**3. **Visit the node.

For the tree in Figure 8.11, visiting the nodes with a postorder traversal would generate the expression

ABC+*

This is called

BC+ means “apply the last operator in the expression, +, to the first and second things.” The first thing is B and the second thing is C, so this gives us (B+C) in infix.

Inserting this in the original expression ABC+* (postfix) gives us A*(B+C) postfix.

**NOTE**

The code in Listing 8.1 contains methods for preorder and postorder traversals, as well as for inorder.

We won’t show the details here, but you can fairly easily construct a tree like that in Figure 8.11 using a postfix expression as input. The approach is analogous to that of evaluating a postfix expression, which we saw in the postfix.java program (Listing 4.8 in Chapter 4). However, instead of storing operands on the stack, we store entire subtrees. We read along the postfix string as we did in postfix.java. Here are the steps when we encounter an operand:

**1. **Make a tree with one node that holds the operand.

**2. **Push this tree onto the stack.

Here are the steps when we encounter an operator:

**1. **Pop two operand trees B and C off the stack.

**2. **Create a new tree A with the operator in its root.

**3. **Attach B as the right child of A.

**4. **Attach C as the left child of A.

**5. **Push the resulting tree back on the stack.

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When you’re done evaluating the postfix string, you pop the one remaining item off the stack. Somewhat amazingly, this item is a complete tree depicting the algebraic expression. You can see the prefix and infix representations of the original postfix (and recover the postfix expression) by traversing the tree as we described. We’ll leave an implementation of this process as an exercise.

**Finding Maximum and Minimum Values**

Incidentally, we should note how easy it is to find the maximum and minimum values in a binary search tree. In fact, this process is so easy we don’t include it as an option in the Workshop applet, nor show code for it in Listing 8.1. Still, understanding how it works is important.

For the minimum, go to the left child of the root; then go to the left child of that child, and so on, until you come to a node that has no left child. This node is the minimum, as shown in Figure 8.12.

63

47

71

Minimum

22

53

67

11

33

50

60

17

49

51

**FIGURE 8.12**

Minimum value of a tree.

Here’s some code that returns the node with the minimum key value: public Node minimum() // returns node with minimum key value

{

Node current, last;

current = root; // start at root

while(current != null) // until the bottom,

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{

last = current; // remember node

current = current.leftChild; // go to left child

}

return last;

}

We’ll need to know about finding the minimum value when we set about deleting a node.

For the **Deleting a Node**

Deleting a node is the most complicated common operation required for binary search trees. However, deletion is important in many tree applications, and studying the details builds character.

You start by finding the node you want to delete, using the same approach we saw in find() and insert(). When you’ve found the node, there are three cases to consider:

**1. **The node to be deleted is a leaf (has no children).

**2. **The node to be deleted has one child.

**3. **The node to be deleted has two children.

We’ll look at these three cases in turn. The first is easy; the second, almost as easy; and the third, quite complicated.

**Case 1: The Node to Be Deleted Has No Children**

To delete a leaf node, you simply change the appropriate child field in the node’s parent to point to null, instead of to the node. The node will still exist, but it will no longer be part of the tree. This is shown in Figure 8.13.

Because of Java’s garbage collection feature, we don’t need to worry about explicitly deleting the node itself. When Java realizes that nothing in the program refers to the node, it will be removed from memory. (In C and C++ you would need to execute free() or delete() to remove the node from memory.)

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10

10

5

5

null

Awaiting

3

7

3

7

garbage

collection

a) Before deletion

b) After deletion

**FIGURE 8.13**

Deleting a node with no children.

**Using the Workshop Applet to Delete a Node with No Children** Assume you’re going to delete node 7 in Figure 8.13. Press the Del button and enter 7 when prompted. Again, the node must be found before it can be deleted.

Repeatedly pressing Del will take you from 10 to 5 to 7. When the node is found, it’s deleted without incident.

**Java Code to Delete a Node with No Children**

The first part of the delete() routine is similar to find() and insert(). It involves finding the node to be deleted. As with insert(), we need to remember the parent of the node to be deleted so we can modify its child fields. If we find the node, we drop out of the while loop with parent containing the node to be deleted. If we can’t find it, we return from delete() with a value of false.

public boolean delete(int key) // delete node with given key

{ // (assumes non-empty list) Node current = root;

Node parent = root;

boolean isLeftChild = true;

while(current.iData != key) // search for node

{

parent = current;

if(key < current.iData) // go left?

{

isLeftChild = true;

current = current.leftChild;

}

else // or go right?

{

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isLeftChild = false;

current = current.rightChild;

}

if(current == null) // end of the line, return false; // didn’t find it

} // end while

// found node to delete

// continues...

}

After we’ve found the node, we check first to verify that it has no children. When this is true, we check the special case of the root. If that’s the node to be deleted, we simply set it to null; this empties the tree. Otherwise, we set the parent’s leftChild or rightChild field to null to disconnect the parent from the node.

// delete() continued...

// if no children, simply delete it

if(current.leftChild==null &&

current.rightChild==null)

{

if(current == root) // if root,

root = null; // tree is empty

else if(isLeftChild)

parent.leftChild = null; // disconnect

else // from parent

parent.rightChild = null;

}

// continues...

**Case 2: The Node to Be Deleted Has One Child**

This second case isn’t so bad either. The node has only two connections: to its parent and to its only child. You want to “snip” the node out of this sequence by connecting its parent directly to its child. This process involves changing the appropriate reference in the parent (leftChild or rightChild) to point to the deleted node’s child.

This situation is shown in Figure 8.14.

**Using the Workshop Applet to Delete a Node with One Child** Let’s assume we’re using the Workshop applet on the tree in Figure 8.14 and deleting node 71, which has a left child but no right child. Press Del and enter 71 when prompted. Keep pressing Del until the arrow rests on 71. Node 71 has only one child, 63. It doesn’t matter whether 63 has children of its own; in this case it has one: 67.

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80

80

52

52

To be

deleted

48

71

48

63

63

67

67

a) Before deletion

b) After deletion

**FIGURE 8.14**

Deleting a node with one child.

Pressing Del once more causes 71 to be deleted. Its place is taken by its left child, 63.

In fact, the entire subtree of which 63 is the root is moved up and plugged in as the new right child of 52.

Use the Workshop applet to generate new trees with one-child nodes, and see what happens when you delete them. Look for the subtree whose root is the deleted node’s child. No matter how complicated this subtree is, it’s simply moved up and plugged in as the new child of the deleted node’s parent.

**Java Code to Delete a Node with One Child**

The following code shows how to deal with the one-child situation. There are four variations: The child of the node to be deleted may be either a left or right child, and for each of these cases the node to be deleted may be either the left or right child of its parent.

There is also a specialized situation: the node to be deleted may be the root, in which case it has no parent and is simply replaced by the appropriate subtree. Here’s the code (which continues from the end of the no-child code fragment shown earlier):

// delete() continued...

// if no right child, replace with left subtree

else if(current.rightChild==null)

if(current == root)

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root = current.leftChild;

else if(isLeftChild) // left child of parent parent.leftChild = current.leftChild;

else // right child of parent parent.rightChild = current.leftChild;

// if no left child, replace with right subtree

else if(current.leftChild==null)

if(current == root)

root = current.rightChild;

else if(isLeftChild) // left child of parent parent.leftChild = current.rightChild;

else // right child of parent parent.rightChild = current.rightChild;

// continued...

Notice that working with references makes it easy to move an entire subtree. You do this by simply disconnecting the old reference to the subtree and creating a new reference to it somewhere else. Although there may be lots of nodes in the subtree, you don’t need to worry about moving them individually. In fact, they “move” only in the sense of being conceptually in different positions relative to the other nodes.

As far as the program is concerned, only the reference to the root of the subtree has changed.

**Case 3: The Node to Be Deleted Has Two Children**

Now the fun begins. If the deleted node has two children, you can’t just replace it with one of these children, at least if the child has its own children. Why not?

Examine Figure 8.15, and imagine deleting node 25 and replacing it with its right subtree, whose root is 35. Which left child would 35 have? The deleted node’s left child, 15, or the new node’s left child, 30? In either case 30 would be in the wrong place, but we can’t just throw it away.

We need another approach. The good news is that there’s a trick. The bad news is that, even with the trick, there are a lot of special cases to consider. Remember that in a binary search tree the nodes are arranged in order of ascending keys. For each node, the node with the next-highest key is called its

Here’s the trick: To delete a node with two children,

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50

Which

50

node goes

here?

To be

Successor

deleted

25

35

30

Root of

right

15

35

15

40

subtree

5

20

30

40

5

20

a) Before deletion

b) After deletion

**FIGURE 8.15**

Cannot replace with subtree.

50

50

To be

deleted

25

30

15

35

15

35

5

20

30

40

5

20

40

Successor

to 25

a) Before deletion

b) After deletion

**FIGURE 8.16**

Node replaced by its successor.

**Finding the Successor**

How do you find the successor of a node? As a human being, you can do this quickly (for small trees, anyway). Just take a quick glance at the tree and find the next-largest number following the key of the node to be deleted. In Figure 8.16 it doesn’t take

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long to see that the successor of 25 is 30. There’s just no other number that is greater than 25 and also smaller than 35. However, the computer can’t do things “at a glance”; it needs an algorithm. Here it is:

First, the program goes to the original node’s right child, which must have a key larger than the node. Then it goes to this right child’s left child (if it has one), and to this left child’s left child, and so on, following down the path of left children. The last left child in this path is the successor of the original node, as shown in Figure 8.17.

To find successor

of this node

38

Go to

right child

26

72

Go to

left child

55

90

Go to

left child

Successor

41

60

78

92

No left

43

74

child

**FIGURE 8.17**

Finding the successor.

Why does this algorithm work? What we’re really looking for is

If the right child of the original node has no left children, this right child is itself the successor, as shown in Figure 8.18.

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To find successor

38

of this node

Go to right child

Successor

72

No left

child

90

78

92

**FIGURE 8.18**

The right child is the successor.

**Using the Workshop Applet to Delete a Node with Two Children** Generate a tree with the Workshop applet, and pick a node with two children. Now mentally figure out which node is its successor, by going to its right child and then following down the line of this right child’s left children (if it has any). You may want to make sure the successor has no children of its own. If it does, the situation gets more complicated because entire subtrees are moved around, rather than a single node.

After you’ve chosen a node to delete, click the Del button. You’ll be asked for the key value of the node to delete. When you’ve specified it, repeated presses of the Del button will show the red arrow searching down the tree to the designated node.

When the node is deleted, it’s replaced by its successor.

Let’s assume you use the Workshop applet to delete the node with key 30 from the example shown earlier in Figure 8.15. The red arrow will go from the root at 50 to 25; then 25 will be replaced by 30.

**Java Code to Find the Successor**

Here’s some code for a method getSuccessor(), which returns the successor of the node specified as its delNode argument. (This routine assumes that delNode does indeed have a right child, but we know this is true because we’ve already determined that the node to be deleted has two children.)

// returns node with next-highest value after delNode

// goes to right child, then right child’s left descendants private node getSuccessor(node delNode)

{

Node successorParent = delNode;

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Node successor = delNode;

Node current = delNode.rightChild; // go to right child while(current != null) // until no more

{ // left children, successorParent = successor;

successor = current;

current = current.leftChild; // go to left child

}

// if successor not

if(successor != delNode.rightChild) // right child,

{ // make connections successorParent.leftChild = successor.rightChild;

successor.rightChild = delNode.rightChild;

}

return successor;

}

The routine first goes to delNode’s right child and then, in the while loop, follows down the path of all this right child’s left children. When the while loop exits, successor contains delNode’s successor.

When we’ve found the successor, we may need to access its parent, so within the while loop we also keep track of the parent of the current node.

The getSuccessor() routine carries out two additional operations in addition to finding the successor. However, to understand them, we need to step back and consider the big picture.

As we’ve seen, the successor node can occupy one of two possible positions relative to current, the node to be deleted. The successor can be current’s right child, or it can be one of this right child’s left descendants. We’ll look at these two situations in turn.

**Successor Is Right Child of **delNode

If successor is the right child of current, things are simplified somewhat because we can simply move the subtree of which successor is the root and plug it in where the deleted node was. This operation requires only two steps: **1. **Unplug current from the rightChild field of its parent (or leftChild field if appropriate), and set this field to point to successor.

**2. **Unplug current’s left child from current, and plug it into the leftChild field of successor.

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Here are the code statements that carry out these steps, excerpted from delete(): **1. **parent.rightChild = successor;

**2. **successor.leftChild = current.leftChild;

This situation is summarized in Figure 8.19, which shows the connections affected by these two steps.

Successor’s

50

Parent

50

parent to be

deleted

Step 1

(“current”)

Step 1

75

87

Step 2

Step 2

62

87

62

93

Successor

Cannot

93

exist

a) Before deletion

b) After deletion

**FIGURE 8.19**

Deletion when the successor is the right child.

Here’s the code in context (a continuation of the else-if ladder shown earlier):

// delete() continued

else // two children, so replace with inorder successor

{

// get successor of node to delete (current)

Node successor = getSuccessor(current);

// connect parent of current to successor instead

if(current == root)

root = successor;

else if(isLeftChild)

parent.leftChild = successor;

else

parent.rightChild = successor;

// connect successor to current’s left child

successor.leftChild = current.leftChild;

} // end else two children

// (successor cannot have a left child)

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return true;

} // end delete()

Notice that this is—finally—the end of the delete() routine. Let’s review the code for these two steps:

• Step 1: If the node to be deleted, current, is the root, it has no parent so we merely set the root to the successor. Otherwise, the node to be deleted can be either a left or right child (Figure 8.19 shows it as a right child), so we set the appropriate field in its parent to point to successor. When delete() returns and current goes out of scope, the node referred to by current will have no references to it, so it will be discarded during Java’s next garbage collection.

• Step 2: We set the left child of successor to point to current’s left child.

What happens if the successor has children of its own? First of all,

Well, remember that the algorithm we use to determine the successor goes to the right child first and then to any left children of that right child. It stops when it gets to a node with no left child, so the algorithm itself determines that the successor can’t have any left children. If it did, that left child would be the successor instead.

You can check this out on the Workshop applet. No matter how many trees you make, you’ll never find a situation in which a node’s successor has a left child (assuming the original node has two children, which is the situation that leads to all this trouble in the first place).

On the other hand, the successor may very well have a right child. This isn’t much of a problem when the successor is the right child of the node to be deleted. When we move the successor, its right subtree simply follows along with it. There’s no conflict with the right child of the node being deleted because the successor

In the next section we’ll see that a successor’s right child needs more attention if the successor is not the right child of the node to be deleted.

**Successor Is Left Descendant of Right Child of **delNode If successor is a left descendant of the right child of the node to be deleted, four steps are required to perform the deletion:

**1. **Plug the right child of successor into the leftChild field of the successor’s parent.

**2. **Plug the right child of the node to be deleted into the rightChild field of successor.

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**3. **Unplug current from the rightChild field of its parent, and set this field to point to successor.

**4. **Unplug current’s left child from current, and plug it into the leftChild field of successor.

Steps 1 and 2 are handled in the getSuccessor() routine, while 3 and 4 are carried out in delete(). Figure 8.20 shows the connections affected by these four steps.

50

Parent

50

Step 3

To be

Step 3

deleted

75

77

Successor’s

Step 4

Step 2

parent

Step 4

Step 2

62

87

62

87

Step 1

Step 1

Successor

77

93

79

93

Step 1

Cannot

Successor’s

79

exist

right child

a) Before deletion

b) After deletion

**FIGURE 8.20**

Deletion when the successor is the left child.

Here’s the code for these four steps:

**1. **successorParent.leftChild = successor.rightChild; **2. **successor.rightChild = delNode.rightChild; **3. **parent.rightChild = successor;

**4. **successor.leftChild = current.leftChild;

(Step 3 could also refer to the left child of its parent.) The numbers in Figure 8.20

show the connections affected by the four steps. Step 1 in effect

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// if successor not

if(successor != delNode.rightChild) // right child,

{ // make connections successorParent.leftChild = successor.rightChild;

successor.rightChild = delNode.rightChild;

}

These steps are more convenient to perform here than in delete(), because in getSuccessor() we can easily figure out where the successor’s parent is while we’re descending the tree to find the successor.

Steps 3 and 4 we’ve seen already; they’re the same as steps 1 and 2 in the case where the successor is the right child of the node to be deleted, and the code is in the if statement at the end of delete().

**Is Deletion Necessary? **

If you’ve come this far, you can see that deletion is fairly involved. In fact, it’s so complicated that some programmers try to sidestep it altogether. They add a new Boolean field to the node class, called something like isDeleted. To delete a node, they simply set this field to true. Then other operations, like find(), check this field to be sure the node isn’t marked as deleted before working with it. This way, deleting a node doesn’t change the structure of the tree. Of course, it also means that memory can fill up with “deleted” nodes.

This approach is a bit of a cop-out, but it may be appropriate where there won’t be many deletions in a tree. (If ex-employees remain in the personnel file forever, for example.)

**The Efficiency of Binary Trees**

As you’ve seen, most operations with trees involve descending the tree from level to level to find a particular node. How long does it take to do this? In a full tree, about half the nodes are on the bottom level. (More accurately, if it’s full, there’s one more node on the bottom row than in the rest of the tree.) Thus, about half of all searches or insertions or deletions require finding a node on the lowest level. (An additional quarter of these operations require finding the node on the next-to-lowest level, and so on.)

During a search we need to visit one node on each level. So we can get a good idea how long it takes to carry out these operations by knowing how many levels there are. Assuming a full tree, Table 8.2 shows how many levels are necessary to hold a given number of nodes.

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**TABLE 8.2**

Number of Levels for Specified Number of Nodes

**Number of Nodes**

**Number of Levels**

1

1

3

2

7

3

15

4

31

5

…

…

1,023

10

…

…

32,767

15

…

…

1,048,575

20

…

…

33,554,432

25

…

…

1,073,741,824

30

This situation is very much like the ordered array discussed in Chapter 2. In that case, the number of comparisons for a binary search was approximately equal to the base 2 logarithm of the number of cells in the array. Here, if we call the number of nodes in the first column N, and the number of levels in the second column L, we can say that N is 1 less than 2 raised to the power L, or N = 2L – 1

Adding 1 to both sides of the equation, we have

N + 1 = 2L

This is equivalent to

L = log (N + 1)

2

Thus, the time needed to carry out the common tree operations is proportional to the base 2 log of N. In Big O notation we say such operations take O(logN) time.

If the tree isn’t full, analysis is difficult. We can say that for a tree with a given number of levels, average search times will be shorter for the non-full tree than the full tree because fewer searches will proceed to lower levels.

Compare the tree to the other data storage structures we’ve discussed so far. In an unordered array or a linked list containing 1,000,000 items, finding the item you want takes, on the average, 500,000 comparisons. But in a tree of 1,000,000 items, only 20 (or fewer) comparisons are required.

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In an ordered array you can find an item equally quickly, but inserting an item requires, on the average, moving 500,000 items. Inserting an item in a tree with 1,000,000 items requires 20 or fewer comparisons, plus a small amount of time to connect the item.

Similarly, deleting an item from a 1,000,000-item array requires moving an average of 500,000 items, while deleting an item from a 1,000,000-node tree requires 20 or fewer comparisons to find the item, plus (possibly) a few more comparisons to find its successor, plus a short time to disconnect the item and connect its successor.

Thus, a tree provides high efficiency for all the common data storage operations.

Traversing is not as fast as the other operations. However, traversals are probably not very commonly carried out in a typical large database. They’re more appropriate when a tree is used as an aid to parsing algebraic or similar expressions, which are probably not too long anyway.

**Trees Represented as Arrays**

Our code examples are based on the idea that a tree’s edges are represented by leftChild and rightChild references in each node. However, there’s a completely different way to represent a tree: with an array.

In the array approach, the nodes are stored in an array and are not linked by references. The position of the node in the array corresponds to its position in the tree.

The node at index 0 is the root, the node at index 1 is the root’s left child, and so on, progressing from left to right along each level of the tree. This is shown in Figure 8.21.

Every position in the tree, whether it represents an existing node or not, corresponds to a cell in the array. Adding a node at a given position in the tree means inserting the node into the equivalent cell in the array. Cells representing tree positions with no nodes are filled with 0 or null.

With this scheme, a node’s children and parent can be found by applying some simple arithmetic to the node’s index number in the array. If a node’s index number is index, this node’s left child is

2*index + 1

its right child is

2*index + 2

and its parent is

(index-1) / 2

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(where the / character indicates integer division with no remainder). You can check this out by looking at Figure 8.21.

Array

50

0

50

0

1

25

2

75

25

75

3

null

1

2

4

37

5

62

6

84

37

62

84

3

4

5

6

7

null

8

null

9

31

31

43

55

92

10

43

7

8

9

10

11

12

13

14

11

55

12

null

13

null

14

92

**FIGURE 8.21**

Tree represented by an array.

In most situations, representing a tree with an array isn’t very efficient. Unfilled nodes and deleted nodes leave holes in the array, wasting memory. Even worse, when deletion of a node involves moving subtrees, every node in the subtree must be moved to its new location in the array, which is time-consuming in large trees.

However, if deletions aren’t allowed, the array representation may be useful, especially if obtaining memory for each node dynamically is, for some reason, too time-consuming. The array representation may also be useful in special situations. The tree in the Workshop applet, for example, is represented internally as an array to make it easy to map the nodes from the array to fixed locations on the screen display.

**Duplicate Keys**

As in other data structures, the problem of duplicate keys must be addressed. In the code shown for insert(), and in the Workshop applet, a node with a duplicate key will be inserted as the right child of its twin.

The problem is that the find() routine will find only the first of two (or more) duplicate nodes. The find() routine could be modified to check an additional data item, to distinguish data items even when the keys were the same, but this would be (at least somewhat) time-consuming.

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One option is to simply forbid duplicate keys. When duplicate keys are excluded by the nature of the data (employee ID numbers, for example), there’s no problem.

Otherwise, you need to modify the insert() routine to check for equality during the insertion process, and abort the insertion if a duplicate is found.

The Fill routine in the Workshop applet excludes duplicates when generating the random keys.

**The Complete **tree.java **Program**

In this section we’ll show the complete program that includes all the methods and code fragments we’ve looked at so far in this chapter. It also features a primitive user interface. This allows the user to choose an operation (finding, inserting, deleting, traversing, and displaying the tree) by entering characters. The display routine uses character output to generate a picture of the tree. Figure 8.22 shows the display generated by the program.

**FIGURE 8.22**

Output of the tree.java program.

In the figure, the user has typed s to display the tree, then typed i and 48 to insert a node with that value, and then s again to display the tree with the additional node.

The 48 appears in the lower display.

The available commands are the characters s, i, f, d, and t, for show, insert, find, delete, and traverse. The i, f, and d options ask for the key value of the node to be operated on. The t option gives you a choice of traversals: 1 for preorder, 2 for inorder, and 3 for postorder. The key values are then displayed in that order.

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The display shows the key values arranged in something of a tree shape; however, you’ll need to imagine the edges. Two dashes (—) represent a node that doesn’t exist at a particular position in the tree. The program initially creates some nodes so the user will have something to see before any insertions are made. You can modify this initialization code to start with any nodes you want, or with no nodes (which is good nodes).

You can experiment with the program in Listing 8.1 as you can with the Workshop applet. Unlike the Workshop applet, however, it doesn’t show you the steps involved in carrying out an operation; it does everything at once.

**LISTING 8.1**

The tree.java Program

// tree.java

// demonstrates binary tree

// to run this program: C>java TreeApp

import java.io.*;

import java.util.*; // for Stack class

////////////////////////////////////////////////////////////////

class Node

{

public int iData; // data item (key)

public double dData; // data item

public Node leftChild; // this node’s left child public Node rightChild; // this node’s right child public void displayNode() // display ourself

{

System.out.print(‘{‘);

System.out.print(iData);

System.out.print(“, “);

System.out.print(dData);

System.out.print(“} “);

}

} // end class Node

////////////////////////////////////////////////////////////////

class Tree

{

private Node root; // first node of tree

// -------------------------------------------------------------

public Tree() // constructor

{ root = null; } // no nodes in tree yet

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**LISTING 8.1**

Continued

// -------------------------------------------------------------

public Node find(int key) // find node with given key

{ // (assumes non-empty tree) Node current = root; // start at root while(current.iData != key) // while no match,

{

if(key < current.iData) // go left?

current = current.leftChild;

else // or go right?

current = current.rightChild;

if(current == null) // if no child,

return null; // didn’t find it

}

return current; // found it

} // end find()

// -------------------------------------------------------------

public void insert(int id, double dd)

{

Node newNode = new Node(); // make new node

newNode.iData = id; // insert data

newNode.dData = dd;

if(root==null) // no node in root

root = newNode;

else // root occupied

{

Node current = root; // start at root

Node parent;

while(true) // (exits internally)

{

parent = current;

if(id < current.iData) // go left?

{

current = current.leftChild;

if(current == null) // if end of the line,

{ // insert on left

parent.leftChild = newNode;

return;

}

} // end if go left

else // or go right?

{

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**LISTING 8.1**

Continued

current = current.rightChild;

if(current == null) // if end of the line

{ // insert on right

parent.rightChild = newNode;

return;

}

} // end else go right

} // end while

} // end else not root

} // end insert()

// -------------------------------------------------------------

public boolean delete(int key) // delete node with given key

{ // (assumes non-empty list) Node current = root;

Node parent = root;

boolean isLeftChild = true;

while(current.iData != key) // search for node

{

parent = current;

if(key < current.iData) // go left?

{

isLeftChild = true;

current = current.leftChild;

}

else // or go right?

{

isLeftChild = false;

current = current.rightChild;

}

if(current == null) // end of the line, return false; // didn’t find it

} // end while

// found node to delete

// if no children, simply delete it

if(current.leftChild==null &&

current.rightChild==null)

{

if(current == root) // if root,

root = null; // tree is empty

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**LISTING 8.1**

Continued

else if(isLeftChild)

parent.leftChild = null; // disconnect

else // from parent

parent.rightChild = null;

}

// if no right child, replace with left subtree

else if(current.rightChild==null)

if(current == root)

root = current.leftChild;

else if(isLeftChild)

parent.leftChild = current.leftChild;

else

parent.rightChild = current.leftChild;

// if no left child, replace with right subtree

else if(current.leftChild==null)

if(current == root)

root = current.rightChild;

else if(isLeftChild)

parent.leftChild = current.rightChild;

else

parent.rightChild = current.rightChild;

else // two children, so replace with inorder successor

{

// get successor of node to delete (current)

Node successor = getSuccessor(current);

// connect parent of current to successor instead

if(current == root)

root = successor;

else if(isLeftChild)

parent.leftChild = successor;

else

parent.rightChild = successor;

// connect successor to current’s left child

successor.leftChild = current.leftChild;

} // end else two children

// (successor cannot have a left child)

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**LISTING 8.1**

Continued

return true; // success

} // end delete()

// -------------------------------------------------------------

// returns node with next-highest value after delNode

// goes to right child, then right child’s left descendents private Node getSuccessor(Node delNode)

{

Node successorParent = delNode;

Node successor = delNode;

Node current = delNode.rightChild; // go to right child while(current != null) // until no more

{ // left children, successorParent = successor;

successor = current;

current = current.leftChild; // go to left child

}

// if successor not

if(successor != delNode.rightChild) // right child,

{ // make connections successorParent.leftChild = successor.rightChild;

successor.rightChild = delNode.rightChild;

}

return successor;

}

// -------------------------------------------------------------

public void traverse(int traverseType)

{

switch(traverseType)

{

case 1: System.out.print(“\nPreorder traversal: “); preOrder(root);

break;

case 2: System.out.print(“\nInorder traversal: “); inOrder(root);

break;

case 3: System.out.print(“\nPostorder traversal: “); postOrder(root);

break;

}

System.out.println();

}

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**LISTING 8.1**

Continued

// -------------------------------------------------------------

private void preOrder(Node localRoot)

{

if(localRoot != null)

{

System.out.print(localRoot.iData + “ “);

preOrder(localRoot.leftChild);

preOrder(localRoot.rightChild);

}

}

// -------------------------------------------------------------

private void inOrder(Node localRoot)

{

if(localRoot != null)

{

inOrder(localRoot.leftChild);

System.out.print(localRoot.iData + “ “);

inOrder(localRoot.rightChild);

}

}

// -------------------------------------------------------------

private void postOrder(Node localRoot)

{

if(localRoot != null)

{

postOrder(localRoot.leftChild);

postOrder(localRoot.rightChild);

System.out.print(localRoot.iData + “ “);

}

}

// -------------------------------------------------------------

public void displayTree()

{

Stack globalStack = new Stack();

globalStack.push(root);

int nBlanks = 32;

boolean isRowEmpty = false;

System.out.println(

“......................................................”); while(isRowEmpty==false)

{

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**CHAPTER 8**

Binary Trees

**LISTING 8.1**

Continued

Stack localStack = new Stack();

isRowEmpty = true;

for(int j=0; j<nBlanks; j++)

System.out.print(‘ ‘);

while(globalStack.isEmpty()==false)

{

Node temp = (Node)globalStack.pop();

if(temp != null)

{

System.out.print(temp.iData);

localStack.push(temp.leftChild);

localStack.push(temp.rightChild);

if(temp.leftChild != null ||

temp.rightChild != null)

isRowEmpty = false;

}

else

{

System.out.print(“--”);

localStack.push(null);

localStack.push(null);

}

for(int j=0; j<nBlanks*2-2; j++)

System.out.print(‘ ‘);

} // end while globalStack not empty

System.out.println();

nBlanks /= 2;

while(localStack.isEmpty()==false)

globalStack.push( localStack.pop() );

} // end while isRowEmpty is false

System.out.println(

“......................................................”);

} // end displayTree()

// -------------------------------------------------------------

} // end class Tree

////////////////////////////////////////////////////////////////

class TreeApp

{

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**LISTING 8.1**

Continued

public static void main(String[] args) throws IOException

{

int value;

Tree theTree = new Tree();

theTree.insert(50, 1.5);

theTree.insert(25, 1.2);

theTree.insert(75, 1.7);

theTree.insert(12, 1.5);

theTree.insert(37, 1.2);

theTree.insert(43, 1.7);

theTree.insert(30, 1.5);

theTree.insert(33, 1.2);

theTree.insert(87, 1.7);

theTree.insert(93, 1.5);

theTree.insert(97, 1.5);

while(true)

{

System.out.print(“Enter first letter of show, “);

System.out.print(“insert, find, delete, or traverse: “); int choice = getChar();

switch(choice)

{

case ‘s’:

theTree.displayTree();

break;

case ‘i’:

System.out.print(“Enter value to insert: “);

value = getInt();

theTree.insert(value, value + 0.9);

break;

case ‘f’:

System.out.print(“Enter value to find: “);

value = getInt();

Node found = theTree.find(value);

if(found != null)

{

System.out.print(“Found: “);

found.displayNode();

System.out.print(“\n”);

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**LISTING 8.1**

Continued

}

else

System.out.print(“Could not find “);

System.out.print(value + ‘\n’);

break;

case ‘d’:

System.out.print(“Enter value to delete: “);

value = getInt();

boolean didDelete = theTree.delete(value);

if(didDelete)

System.out.print(“Deleted “ + value + ‘\n’);

else

System.out.print(“Could not delete “);

System.out.print(value + ‘\n’);

break;

case ‘t’:

System.out.print(“Enter type 1, 2 or 3: “);

value = getInt();

theTree.traverse(value);

break;

default:

System.out.print(“Invalid entry\n”);

} // end switch

} // end while

} // end main()

// -------------------------------------------------------------

public static String getString() throws IOException

{

String s = br.readLine();

return s;

}

// -------------------------------------------------------------

public static char getChar() throws IOException

{

String s = getString();

return s.charAt(0);

}

//-------------------------------------------------------------

public static int getInt() throws IOException

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**LISTING 8.1**

Continued

{

String s = getString();

return Integer.parseInt(s);

}

// -------------------------------------------------------------

} // end class TreeApp

////////////////////////////////////////////////////////////////

You can use the ç+C key combination to exit from this program; in the interest of simplicity there’s no single-letter key for this action.

**The Huffman Code**

You shouldn’t get the idea that binary trees are always search trees. Many binary trees are used in other ways. We saw an example in Figure 8.11, where a binary tree represents an algebraic expression.

In this section we’ll discuss an algorithm that uses a binary tree in a surprising way to compress data. It’s called the Huffman code, after David Huffman who discovered it in 1952. Data compression is important in many situations. An example is sending data over the Internet, where, especially over a dial-up connection, transmission can take a long time. An implementation of this scheme is somewhat lengthy, so we won’t show a complete program. Instead, we’ll focus on the concepts and leave the implementation as an exercise.

**Character Codes**

Each character in a normal uncompressed text file is represented in the computer by one byte (for the venerable ASCII code) or by two bytes (for the newer Unicode, which is designed to work for all languages.) In these schemes, every character requires the same number of bits. Table 8.3 shows how some characters are represented in binary using the ASCII code. As you can see, every character takes 8 bits.

**TABLE 8.3**

Some ASCII Codes

**Character**

**Decimal**

**Binary**

A

65

01000000

B

66

01000001

C

67

01000010

…

…

…

X

88

01011000

Y

89

01011001

Z

90

01011010

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There are several approaches to compressing data. For text, the most common approach is to reduce the number of bits that represent the most-used characters. In English, E is often the most common letter, so it seems reasonable to use as few bits as possible to encode it. On the other hand, Z is seldom used, so using a large number of bits is not so bad.

Suppose we use just two bits for E, say 01. We can’t encode every letter of the alphabet in two bits because there are only four 2-bit combinations: 00, 01, 10, and 11.

Can we use these four combinations for the four most-used characters?

Unfortunately not. We must be careful that no character is represented by the same bit combination that appears at the beginning of a longer code used for some other character. For example, if E is 01, and X is 01011000, then anyone decoding 01011000 wouldn’t know if the initial 01 represented an E or the beginning of an X.

This leads to a rule:

Something else to consider is that in some messages E might not be the most-used character. If the text is a Java source file, for example, the ; (semicolon) character might appear more often than E. Here’s the solution to that problem: For each message, we make up a new code tailored to that particular message. Suppose we want to send the message SUSIE SAYS IT IS EASY. The letter S appears a lot, and so does the space character. We might want to make up a table showing how many times each letter appears. This is called a frequency table, as shown in Table 8.4.

**TABLE 8.4**

Frequency Table

**Character**

**Count**

A

2

E

2

I

3

S

6

T

1

U

1

Y

2

Space

4

Linefeed

1

The characters with the highest counts should be coded with a small number of bits.

Table 8.5 shows how we might encode the characters in the Susie message.

**TABLE 8.5**

Huffman Code

**Character**

**Code**

A

010

E

1111

I

110

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**TABLE 8.5**

Continued

**Character**

**Code**

S

10

T

0110

U

01111

Y

1110

Space

00

Linefeed

01110

We use 10 for S and 00 for the space. We can’t use 01 or 11 because they are prefixes for other characters. What about 3-bit combinations? There are eight possibilities: 000, 001, 010, 011, 100, 101, 110, and 111. A is 010 and I is 110. Why aren’t any other combinations used? We already know we can’t use anything starting with 10

or 00; that eliminates four possibilities. Also, 011 is used at the beginning of U and the linefeed, and 111 is used at the beginning of E and Y. Only two 3-bit codes remain, which we use for A and I. In a similar way we can see why only three 4-bit codes are available.

Thus, the entire message is coded as

10 01111 10 110 1111 00 10 010 1110 10 00 110 0110 00 110 10 00

➥ 1111 010 10 1110 01110

For sanity reasons we show this message broken into the codes for individual characters. Of course, in reality all the bits would run together; there is no space character in a binary message, only 0s and 1s.

**Decoding with the Huffman Tree**

We’ll see later how to create Huffman codes. First, we’ll examine the somewhat easier process of decoding. Suppose we received the string of bits shown in the preceding section. How would we transform it back into characters? We can use a kind of binary tree called a

The characters in the message appear in the tree as leaf nodes. The higher their frequency in the message, the higher up they appear in the tree. The number outside each circle is the frequency. The numbers outside non-leaf nodes are the sums of the frequencies of their children. We’ll see later why this is important.

How do we use this tree to decode the message? For each character you start at the root. If you see a 0 bit, you go left to the next node, and if you see a 1 bit, you go right. Try it with the code for A, which is 010. You go left, then right, then left again, and,

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22

0

9

13

4

1

5

6

7

sp

S

0

2

3

3

4

A

I

1

2

2

2

T

Y

E

1

1

lf

U

**FIGURE 8.23**

Huffman tree.

You’ll see you can do the same with the other characters. If you have the patience, you can decode the entire bit string this way.

**Creating the Huffman Tree**

We’ve seen how to use the Huffman tree for decoding, but how do we create this tree? There are many ways to handle this problem. We’ll base our approach on the Node and Tree classes in the tree.java program in Listing 8.1 (although routines that are specific to search trees, like find(), insert(), and delete() are no longer relevant).

Here is the algorithm for constructing the tree:

**1. **Make a Node object (as seen in tree.java) for each character used in the message. For our Susie example that would be nine nodes. Each node has two data items: the character and that character’s frequency in the message. Table 8.4 provides this information for the Susie message.

**2. **Make a tree object for each of these nodes. The node becomes the root of the tree.

**3. **Insert these trees in a priority queue (as described in Chapter 4). They are ordered by frequency, with the smallest frequency having the highest priority.

That is, when you remove a tree, it’s always the one with the least-used character.

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Now do the following:

**1. **Remove two trees from the priority queue, and make them into children of a new node. The new node has a frequency that is the sum of the children’s frequencies; its character field can be left blank.

**2. **Insert this new three-node tree back into the priority queue.

**3. **Keep repeating steps 1 and 2. The trees will get larger and larger, and there will be fewer and fewer of them. When there is only one tree left in the queue, it is the Huffman tree and you’re done.

Figures 8.24 and 8.25 show how the Huffman tree is constructed for the Susie message.

1

1

1

2

2

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**FIGURE 8.24**

Growing the Huffman tree, Part 1.

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Growing the Huffman tree, Part 2.

**Coding the Message**

Now that we have the Huffman tree, how do we code a message? We start by creating a code table, which lists the Huffman code alongside each character. To simplify the discussion, let’s assume that, instead of the ASCII code, our computer uses a simplified alphabet that has only uppercase letters with 28 characters. A is 0, B is 1, and so on up to Z, which is 25. A space is 26, and a linefeed is 27. We number these characters so their numerical codes run from 0 to 27. (This is not a compressed code, just a simplification of the ASCII code, the normal way characters are stored in the computer.)

Our code table would be an array of 28 cells. The index of each cell would be the numerical value of the character: 0 for A, 1 for B, and so on. The contents of the cell would be the Huffman code for the corresponding character. Not every cell contains

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a code; only those that appear in the message. Figure 8.26 shows how this looks for the Susie message.

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linefeed

**FIGURE 8.26**

Code table.

Such a code table makes it easy to generate the coded message: For each character in the original message, we use its code as an index into the code table. We then repeatedly append the Huffman codes to the end of the coded message until it’s complete.

**Creating the Huffman Code**

How do we create the Huffman code to put into the code table? The process is like decoding a message. We start at the root of the Huffman tree and follow every possible path to a leaf node. As we go along the path, we remember the sequence of left and right choices, recording a 0 for a left edge and a 1 for a right edge. When we arrive at the leaf node for a character, the sequence of 0s and 1s is the Huffman code for that character. We put this code into the code table at the appropriate index number.

This process can be handled by calling a method that starts at the root and then calls itself recursively for each child. Eventually, the paths to all the leaf nodes will be explored and the code table will be complete.

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**Summary**

• Trees consist of nodes (circles) connected by edges (lines).

• The root is the topmost node in a tree; it has no parent.

• In a binary tree, a node has at most two children.

• In a binary search tree, all the nodes that are left descendants of node A have key values less than A; all the nodes that are A’s right descendants have key values greater than (or equal to) A.

• Trees perform searches, insertions, and deletions in O(log N) time.

• Nodes represent the data objects being stored in the tree.

• Edges are most commonly represented in a program by references to a node’s children (and sometimes to its parent).

• Traversing a tree means visiting all its nodes in some order.

• The simplest traversals are preorder, inorder, and postorder.

• An unbalanced tree is one whose root has many more left descendents than right descendants, or vice versa.

• Searching for a node involves comparing the value to be found with the key value of a node, and going to that node’s left child if the key search value is less, or to the node’s right child if the search value is greater.

• Insertion involves finding the place to insert the new node and then changing a child field in its new parent to refer to it.

• An inorder traversal visits nodes in order of ascending keys.

• Preorder and postorder traversals are useful for parsing algebraic expressions.

• When a node has no children, it can be deleted by setting the child field in its parent to null.

• When a node has one child, it can be deleted by setting the child field in its parent to point to its child.

• When a node has two children, it can be deleted by replacing it with its successor.

• The successor to a node A can be found by finding the minimum node in the subtree whose root is A’s right child.

• In a deletion of a node with two children, different situations arise, depending on whether the successor is the right child of the node to be deleted or one of the right child’s left descendants.

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• Nodes with duplicate key values may cause trouble in arrays because only the first one can be found in a search.

• Trees can be represented in the computer’s memory as an array, although the reference-based approach is more common.

• A Huffman tree is a binary tree (but not a search tree) used in a data-compression algorithm called Huffman Coding.

• In the Huffman code the characters that appear most frequently are coded with the fewest bits, and those that appear rarely are coded with the most bits.

**Questions**

These questions are intended as a self-test for readers. Answers may be found in Appendix C.

**1. **Insertion and deletion in a tree require what big O time?

**2. **A binary tree is a search tree if

**a. **every non-leaf node has children whose key values are less than (or equal to) the parent.

**b. **every left child has a key less than the parent and every right child has a key greater than (or equal to) the parent.

**c. **in the path from the root to every leaf node, the key of each node is greater than (or equal to) the key of its parent.

**d. **a node can have a maximum of two children.

**3. **True or False: Not all trees are binary trees.

**4. **In a complete binary tree with 20 nodes, and the root considered to be at level 0, how many nodes are there at level 4?

**5. **A subtree of a binary tree always has

**a. **a root that is a child of the main tree’s root.

**b. **a root unconnected to the main tree’s root.

**c. **fewer nodes than the main tree.

**d. **a sibling with the same number of nodes.

**6. **In the Java code for a tree, the ______ and the _______ are generally separate classes.

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**7. **Finding a node in a binary search tree involves going from node to node, asking

**a. **how big the node’s key is in relation to the search key.

**b. **how big the node’s key is compared to its right or left children.

**c. **what leaf node we want to reach.

**d. **what level we are on.

**8. **An unbalanced tree is one

**a. **in which most of the keys have values greater than the average.

**b. **whose behavior is unpredictable.

**c. **in which the root or some other node has many more left children than right children, or vice versa.

**d. **that is shaped like an umbrella.

**9. **Inserting a node starts with the same steps as _______ a node.

**10. **Suppose a node A has a successor node S. Then S must have a key that is larger than _____ but smaller than or equal to _______.

**11. **In a binary tree used to represent a mathematical expression, which of the following is not true?

**a. **Both children of an operator node must be operands.

**b. **Following a postorder traversal, no parentheses need to be added.

**c. **Following an inorder traversal, parentheses must be added.

**d. **In pre-order traversal a node is visited before either of its children.

**12. **If a tree is represented by an array, the right child of a node at index n has an index of _______ .

**13. **True or False: Deleting a node with one child from a binary search tree involves finding that node’s successor.

**14. **A Huffman tree is typically used to _______ text.

**15. **Which of the following is not true about a Huffman tree?

**a. **The most frequently used characters always appear near the top of the tree.

**b. **Normally, decoding a message involves repeatedly following a path from the root to a leaf.

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**c. **In coding a character you typically start at a leaf and work upward.

**d. **The tree can be generated by removal and insertion operations on a priority queue.

**Experiments**

**1. **Use the Binary Tree Workshop applet to create 20 trees. What percentage would you say are seriously unbalanced?

**2. **Create a UML activity diagram (or flowchart, for you old-timers) of the various possibilities when deleting a node from a binary search tree. It should detail the three cases described in the text. Include the variations for left and right children and special cases like deletion of the root. For example, there are two possibilities for case 1 (left and right children). Boxes at the end of each path should describe how to do the deletion in that situation.

**3. **Use the Binary Tree Workshop applet to delete a node in every possible situation.

**Programming Projects**

Writing programs to solve the Programming Projects helps to solidify your understanding of the material and demonstrates how the chapter’s concepts are applied.

(As noted in the Introduction, qualified instructors may obtain completed solutions to the Programming Projects on the publisher’s Web site.) **8.1 **Start with the tree.java program (Listing 8.1) and modify it to create a binary tree from a string of letters (like A, B, and so on) entered by the user. Each letter will be displayed in its own node. Construct the tree so that all the nodes that contain letters are leaves. Parent nodes can contain some non-letter symbol like +. Make sure that every parent node has exactly two children.

Don’t worry if the tree is unbalanced. Note that this will not be a search tree; there’s no quick way to find a given node. You may end up with something like this:

+

+ E

+ D - -

+ C - - - - - -

A B - - - - - - - - - - - - - -

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One way to begin is by making an array of trees. (A group of unconnected trees is called a

The routines find(), insert(), and delete(), which apply only to search trees, can be deleted. Keep the displayTree() method and the traversals because they will work on any binary tree.

**8.2 **Expand the program in Programming Project 8.1 to create a balanced tree. One way to do this is to make sure that as many leaves as possible appear in the bottom row. You can start by making a three-node tree out of each pair of one-node trees, making a new + node for the root. This results in a forest of three-node trees. Then combine each pair of three-node trees to make a forest of seven-node trees. As the number of nodes per tree grows, the number of trees shrinks, until finally there is only one tree left.

**8.3 **Again, start with the tree.java program and make a tree from characters typed by the user. This time, make a complete tree—one that is completely full except possibly on the right end of the bottom row. The characters should be ordered from the top down and from left to right along each row, as if writing a letter on a pyramid. (This arrangement does not correspond to any of the three traversals we discussed in this chapter.) Thus, the string ABCDEFGHIJ

would be arranged as

A

B C

D E F G

H I J

One way to create this tree is from the top down, rather than the bottom up as in the previous two Programming Projects. Start by creating a node which will be the root of the final tree. If you think of the nodes as being numbered in the same order the letters are arranged, with 1 at the root, then any node numbered n has a left child numbered 2*n and a right child numbered 2*n+1.

You might use a recursive routine that makes two children and then calls itself for each child. The nodes don’t need to be created in the same order they are arranged on the tree. As in the previous Programming Projects, you can jettison the search-tree routines from the Tree class.

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**8.4 **Write a program that transforms a postfix expression into a tree such as that shown in Figure 8.11 in this chapter. You’ll need to modify the Tree class from the tree.java program (Listing 8.1) and the ParsePost class from the postfix.java program (Listing 4.8) in Chapter 4. There are more details in the discussion of Figure 8.11.

After the tree is generated, traversals of the tree will yield the prefix, infix, and postfix equivalents of the algebraic expression. The infix version will need parentheses to avoid generating ambiguous expressions. In the inOrder() method, display an opening parenthesis before the first recursive call and a closing parenthesis after the second recursive call.

**8.5 **Write a program to implement Huffman coding and decoding. It should do the following:

Accept a text message, possibly of more than one line.

Create a Huffman tree for this message.

Create a code table.

Encode the message into binary.

Decode the message from binary back to text.

If the message is short, the program should be able to display the Huffman tree after creating it. The ideas in Programming Projects 8.1, 8.2, and 8.3 might prove helpful. You can use String variables to store binary numbers as arrangements of the characters 1 and 0. Don’t worry about doing actual bit manipulation unless you really want to.

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**9**

**IN THIS CHAPTER**

• Our Approach to the

Red-Black Trees

Discussion

• Balanced and Unbalanced

Trees

As you learned in Chapter 8, “Binary Trees,” ordinary

• Experimenting with the

binary search trees offer important advantages as data Workshop Applet

storage devices: You can quickly search for an item with a given key, and you can also quickly insert or delete an

• Rotations

item. Other data storage structures, such as arrays, sorted

• Inserting a New Node

arrays, and linked lists, perform one or the other of these activities slowly. Thus, binary search trees might appear to

• Deletion

be the ideal data storage structure.

• The Efficiency of Red-Black

Unfortunately, ordinary binary search trees suffer from a Trees

troublesome problem. They work well if the data is

•

inserted into the tree in random order. However, they Red-Black Tree

become much slower if data is inserted in already-sorted Implementation

order (17, 21, 28, 36,…) or inversely sorted order (36, 28,

• Other Balanced Trees

21, 17,…). When the values to be inserted are already ordered, a binary tree becomes unbalanced. With an

unbalanced tree, the ability to quickly find (or insert or delete) a given element is lost.

This chapter explores one way to solve the problem of unbalanced trees: the red-black tree, which is a binary search tree with some added features.

There are other ways to ensure that trees are balanced.

We’ll mention some at the end of this chapter, and

examine several, 2-3-4 trees and 2-3 trees, in Chapter 10,

“2-3-4 Trees and External Storage.” In fact, as we’ll see in that chapter, operations on a 2-3-4 tree correspond in a surprising way to operations on a red-black tree.

**Our Approach to the Discussion**

We’ll explain insertion into red-black trees a little differently than we have explained insertion into other data structures. Red-black trees are not trivial to understand.

Because of this and also because of a multiplicity of

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symmetrical cases (for left or right children, inside or outside grandchildren, and so on), the actual code is more lengthy and complex than one might expect. It’s therefore hard to learn about the algorithm by examining code. Accordingly, there are no listings in this chapter. You can create similar functionality using a 2-3-4 tree with the code shown in Chapter 10. However, the concepts you learn about here will aid your understanding of 2-3-4 trees and are themselves quite interesting.

**Conceptual**

For our conceptual understanding of red-black trees, we will be aided by the RBTree Workshop applet. We’ll describe how you can work in partnership with the applet to insert new nodes into a tree. Including a human into the insertion routine certainly slows it down but also makes it easier for the human to understand how the process works.

Searching works the same way in a red-black tree as it does in an ordinary binary tree. On the other hand, insertion and deletion, while based on the algorithms in an ordinary tree, are extensively modified. Accordingly, in this chapter we’ll be concentrating on the insertion process.

**Top-Down Insertion**

The approach to insertion that we’ll discuss is called

Another approach is

Bottom-up insertion is less efficient because two passes must be made through the tree.

**Balanced and Unbalanced Trees**

Before we begin our investigation of red-black trees, let’s review how trees become unbalanced. Fire up the Binary Tree Workshop applet from Chapter 8 (not this chapter’s RBTree applet). Use the Fill button to create a tree with only one node.

Then insert a series of nodes whose keys are in either ascending or descending order.

The result will be something like that in Figure 9.1.

The nodes arrange themselves in a line with no branches. Because each node is larger than the previously inserted one, every node is a right child, so all the nodes are on one side of the root. The tree is maximally unbalanced. If you inserted items in descending order, every node would be the left child of its parent, and the tree would be unbalanced on the other side.

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10

20

30

40

50

**FIGURE 9.1**

Items inserted in ascending order.

**Degenerates to O(N)**

When there are no branches, the tree becomes, in effect, a linked list. The arrangement of data is one-dimensional instead of two-dimensional. Unfortunately, as with a linked list, you must now search through (on the average) half the items to find the one you’re looking for. In this situation the speed of searching is reduced to O(N), instead of O(logN) as it is for a balanced tree. Searching through 10,000 items in such an unbalanced tree would require an average of 5,000 comparisons, whereas for a balanced tree with random insertions it requires only 14. For presorted data you might just as well use a linked list in the first place.

Data that’s only partly sorted will generate trees that are only partly unbalanced. If you use the Binary Tree Workshop applet from Chapter 8 to attempt to generate trees with 31 nodes, you’ll see that some of them are more unbalanced than others, as shown in Figure 9.2.

Although not as bad as a maximally unbalanced tree, this situation is not optimal for searching times.

In the Binary Tree Workshop applet, trees can become partially unbalanced, even with randomly generated data, because the amount of data is so small that even a short run of ordered numbers will have a big effect on the tree. Also, a very small or very large key value can cause an unbalanced tree by not allowing the insertion of many nodes on one side or the other of its node. A root of 3, for example, allows only two more nodes to be inserted to its left.

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**FIGURE 9.2**

A partially unbalanced tree.

With a realistic amount of random data, it’s not likely a tree would become seriously unbalanced. However, there may be runs of sorted data that will partially unbalance a tree. Searching partially unbalanced trees will take time somewhere between O(N) and O(logN), depending on how badly the tree is unbalanced.

**Balance to the Rescue**

To guarantee the quick O(log N) search times a tree is capable of, we need to ensure that our tree is always balanced (or at least almost balanced). This means that each node in a tree needs to have roughly the same number of descendents on its left side as it has on its right.

In a red-black tree, balance is achieved during insertion (and also deletion, but we’ll ignore that for the moment). As an item is being inserted, the insertion routine checks that certain characteristics of the tree are not violated. If they are, it takes corrective action, restructuring the tree as necessary. By maintaining these characteristics, the tree is kept balanced.

**Red-Black Tree Characteristics**

What are these mysterious tree characteristics? There are two, one simple and one more complicated:

• The nodes are colored.

• During insertion and deletion, rules are followed that preserve various arrangements of these colors.

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**Colored Nodes**

In a red-black tree, every node is either black or red. These are arbitrary colors; blue and yellow would do just as well. In fact, the whole concept of saying that nodes have “colors” is somewhat arbitrary. Some other analogy could have been used instead: We could say that every node is either heavy or light, or yin or yang.

However, colors are convenient labels. A data field, which can be boolean (isRed, for example), is added to the node class to embody this color information.

In the RBTree Workshop applet, the red-black characteristic of a node is shown by its border color. The center color, as it was in the Binary Tree Workshop applet in the preceding chapter, is simply a randomly generated data field of the node.

When we speak of a node’s color in this chapter, we’ll almost always be referring to its red-black border color. In the figures (except the screenshot of Figure 9.3) we’ll show black nodes with a solid black border and red nodes with a white border.

(Nodes are sometimes shown with no border to indicate that it doesn’t matter whether they’re black or red.)

**Red-Black Rules**

When inserting (or deleting) a new node, certain rules, which we call the

**1. **Every node is either red or black.

**2. **The root is always black.

**3. **If a node is red, its children must be black (although the converse isn’t necessarily true).

**4. **Every path from the root to a leaf, or to a null child, must contain the same number of black nodes.

The “null child” referred to in Rule 4 is a place where a child could be attached to a non-leaf node. In other words, it’s the potential left child of a node with a right child, or the potential right child of a node with a left child. This will make more sense as we go along.

The number of black nodes on a path from root to leaf is called the

Another way to state Rule 4 is that the black height must be the same for all paths from the root to a leaf.

These rules probably seem completely mysterious. It’s not obvious how they will lead to a balanced tree, but they do; some very clever people invented them. Copy them onto a sticky note, and keep it on your computer. You’ll need to refer to them often in the course of this chapter.

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You can see how the rules work by using the RBTree Workshop applet. We’ll do some experiments with the applet in a moment, but first you should understand what actions you can take to fix things if one of the red-black rules is broken.

**Duplicate Keys**

What happens if there’s more than one data item with the same key? This presents a slight problem in red-black trees. It’s important that nodes with the same key are distributed on both sides of other nodes with the same key. That is, if keys arrive in the order 50, 50, 50, you want the second 50 to go to the right of the first one, and the third 50 to go to the left of the first one. Otherwise, the tree becomes unbalanced.

Distributing nodes with equal keys could be handled by some kind of randomizing process in the insertion algorithm. However, the search process then becomes more complicated if all items with the same key must be found.

It’s simpler to outlaw items with the same key. In this discussion we’ll assume duplicates aren’t allowed.

**Fixing Rule Violations**

Suppose you see (or are told by the applet) that the color rules are violated. How can you fix things so your tree is in compliance? There are two, and only two, possible actions you can take:

• You can change the colors of nodes.

• You can perform rotations.

In the applet, changing the color of a node means changing its red-black border color (not the center color). A rotation is a rearrangement of the nodes that, one hopes, leaves the tree more balanced.

At this point such concepts probably seem very abstract, so let’s become familiar with the RBTree Workshop applet, which can help to clarify things.

**Using the RBTree Workshop Applet**

Figure 9.3 shows what the RBTree Workshop applet looks like after some nodes have been inserted. (It may be hard to tell the difference between red and black node borders in the figure, but they should be clear on a color monitor.) There are quite a few buttons in the RBTree applet. We’ll briefly review what they do, although at this point some of the descriptions may be a bit puzzling.

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**FIGURE 9.3**

The RBTree Workshop applet.

**Clicking on a Node**

The red arrow points to the currently selected node. It’s this node whose color is changed or which is the top node in a rotation. You select a node by single-clicking it with the mouse, which moves the red arrow to the node.

**The Start Button**

When you first start the RBTree Workshop applet, and also when you press the Start button, you’ll see that a tree with only one node is created. Because an understanding of red-black trees focuses on using the red-black rules during the insertion process, it’s more convenient to begin with the root and build up the tree by inserting additional nodes. To simplify future operations, the initial root node is always given a value of 50. You select your own numbers for subsequent insertions.

**The Ins Button**

The Ins button causes a new node to be created, with the value that was typed into the Number box, and then inserted into the tree. (At least this is what happens if no color flips are necessary. See the section on the Flip button for more on this possibility.)

Notice that the Ins button does a complete insertion operation with one push; multiple pushes are not required as they were with the Binary Tree Workshop applet in the preceding chapter. It’s therefore important to type the key value before pushing the button. The focus in the RBTree applet is not on the process of finding the place

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to insert the node, which is similar to that in ordinary binary search trees, but on keeping the tree balanced; so the applet doesn’t show the individual steps in the insertion.

**The Del Button**

Pushing the Del button causes the node with the key value typed into the Number box to be deleted. As with the Ins button, this deletion takes place immediately after the first push; multiple pushes are not required.

The Del button and the Ins button use the basic insertion algorithms—the same as those in the Tree Workshop applet. This is how the work is divided between the applet and the user: The applet does the insertion, but it’s (mostly) up to the user to make the appropriate changes to the tree to ensure the red-black rules are followed and the tree thereby becomes balanced.

**The Flip Button**

If there is a black parent with two red children, and you place the red arrow on the parent by clicking on the node with the mouse, then when you press the Flip button, the parent will become red and the children will become black. That is, the colors are flipped between the parent and children. You’ll learn later why such a color exchange is desirable.

If you try to flip the root, it will remain black, so as not to violate Rule 2, but its children will change from red to black.

**The RoL Button**

The RoL button carries out a left rotation. To rotate a group of nodes, first single-click the mouse to position the arrow at the topmost node of the group to be rotated. For a left rotation, the top node must have a right child. Then click the button. We’ll examine rotations in detail later.

**The RoR Button**

The RoR button performs a right rotation. Position the arrow on the top node to be rotated, making sure it has a left child; then click the button.

**The R/B Button**

The R/B button changes a red node to black, or a black node to red. Single-click the mouse to position the red arrow on the node, and then push the button. (This button changes the color of a single node; don’t confuse it with the Flip button, which changes three nodes at once.)

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**Text Messages**

Messages in the text box below the buttons tell you whether the tree is

If it’s not correct, you’ll see messages advising which rule is being violated. In some cases the red arrow will point to the place where the violation occurred.

**Where’s the Find Button? **

In red-black trees, a search routine operates exactly as it did in the ordinary binary search trees described in the preceding chapter. It starts at the root, and, at each node it encounters (the current node), it decides whether to go to the left or right child by comparing the key of the current node with the search key.

We don’t include a Find button in the RBTree applet because you already understand this process and our attention will be on manipulating the red-black aspects of the tree.

**Experimenting with the Workshop Applet**

Now that you’re familiar with the RBTree buttons, let’s do some simple experiments to get a feel for what the applet does. The idea here is to learn to manipulate the applet’s controls. Later you’ll use these skills to balance the tree.

**Experiment 1: Inserting Two Red Nodes**

Press Start to clear any extra nodes. You’ll be left with the root node, which always has the value 50.

Insert a new node with a value smaller than the root, say 25, by typing the number into the Number box and pressing the Ins button. Adding this node doesn’t cause any rule violations, so the message continues to say Tree is red-black correct.

Insert a second node that’s larger than the root, say 75. The tree is still red-black correct. It’s also balanced; there are the same number of nodes on the right of the only non-leaf node (the root) as there are on its left. The result is shown in Figure 9.4.

Notice that newly inserted nodes are always colored red (except for the root). This is not an accident. Inserting a red node is less likely to violate the red-black rules than inserting a black one. This is because, if the new red node is attached to a black one, no rule is broken. It doesn’t create a situation in which there are two red nodes together (Rule 3), and it doesn’t change the black height in any of the paths (Rule 4).

Of course, if you attach a new red node to a red node, Rule 3 will be violated.

However, with any luck this will happen only half the time. Whereas, if it were possible to add a new black node, it would always change the black height for its path, violating Rule 4.

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Black node

50

Red node

Red node

25

75

**FIGURE 9.4**

A balanced tree.

Also, it’s easier to fix violations of Rule 3 (parent and child are both red) than Rule 4

(black heights differ), as we’ll see later.

**Experiment 2: Rotations**

Let’s try some rotations. Start with the three nodes as shown in Figure 9.4. Position the red arrow on the root (50) by clicking it with the mouse. This node will be the

The nodes all shift to new positions, as shown in Figure 9.5.

Arrow

25

50

75

**FIGURE 9.5**

Following a right rotation.

In this right rotation, the parent or top node moves into the place of its right child, the left child moves up and takes the place of the parent, and the right child moves down to become the grandchild of the new top node.

Notice that the tree is now unbalanced; more nodes appear to the right of the root than to the left. Also, the message indicates that the red-black rules are violated, specifically Rule 2 (the root is always black). Don’t worry about this problem yet.

Instead, rotate the other way. Position the red arrow on 25, which is now the root (the arrow should already point to 25 after the previous rotation). Click the RoL

button to rotate left. The nodes will return to the position of Figure 9.4.

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**Experiment 3: Color Flips**

Start with the position of Figure 9.4, with nodes 25 and 75 inserted in addition to 50

in the root position. Note that the parent (the root) is black and both its children are red. Now try to insert another node. No matter what value you use, you’ll see the message Can’t Insert: Needs color flip.

As we mentioned, a color flip is necessary whenever, during the insertion process, a black node with two red children is encountered.

The red arrow should already be positioned on the black parent (the root node), so click the Flip button. The root’s two children change from red to black. Ordinarily, the parent would change from black to red, but this is a special case because it’s the root: It remains black to avoid violating Rule 2. Now all three nodes are black. The tree is still red-black correct.

Now click the Ins button again to insert the new node. Figure 9.6 shows the result if the newly inserted node has the key value 12.

50

25

75

12

**FIGURE 9.6**

Colors flipped, new node inserted.

The tree is still red-black correct. The root is black, there’s no situation in which a parent and child are both red, and all the paths have the same number of black nodes (two). Adding the new red node didn’t change the red-black correctness.

**Experiment 4: An Unbalanced Tree**

Now let’s see what happens when you try to do something that leads to an unbalanced tree. In Figure 9.6 one path has one more node than the other. This isn’t very unbalanced, and no red-black rules are violated, so neither we nor the red-black algorithms need to worry about it. However, suppose that one path differs from another by two or more levels (where level is the same as the number of nodes along the path). In this case the red-black rules will always be violated, and we’ll need to rebalance the tree.

Insert a 6 into the tree of Figure 9.6. You’ll see the message Error: parent and child are both red. Rule 3 has been violated, as shown in Figure 9.7.

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50

25

75

Red parent and

12

child violates

Rule 3

Changing this to

6

black violates

Rule 4

**FIGURE 9.7**

Parent and child are both red.

How can we fix the tree so Rule 3 isn’t violated? An obvious approach is to change one of the offending nodes to black. Let’s try changing the child node, 6. Position the red arrow on it and press the R/B button. The node becomes black.

The good news is we fixed the problem of both parent and child being red. The bad news is that now the message says Error: Black heights differ. The path from the root to node 6 has three black nodes in it, while the path from the root to node 75

has only two. Thus, Rule 4 is violated. It seems we can’t win.

This problem can be fixed with a rotation and some color changes. How to do this will be the topic of later sections.

**More Experiments**

Experiment with the RBTree Workshop applet on your own. Insert more nodes and see what happens. See if you can use rotations and color changes to achieve a balanced tree. Does keeping the tree red-black correct seem to guarantee an (almost) balanced tree?

Try inserting ascending keys (50, 60, 70, 80, 90) and then restart with the Start button and try descending keys (50, 40, 30, 20, 10). Ignore the messages; we’ll see what they mean later. These are the situations that get the ordinary binary search tree into trouble. Can you still balance the tree?

**The Red-Black Rules and Balanced Trees**

Try to create a tree that is unbalanced by two or more levels but is red-black correct.

As it turns out, this is impossible. That’s why the red-black rules keep the tree balanced. If one path is more than one node longer than another, it must either

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have more black nodes, violating Rule 4, or it must have two adjacent red nodes, violating Rule 3. Convince yourself that this is true by experimenting with the applet.

**Null Children**

Remember that Rule 4 specifies all paths that go from the root to any leaf or to

50

25

75

Black height is 2

12

Null child of 25

Black height is 1

Null child of 12

6

Black height is 2

Black height is 2

**FIGURE 9.8**

Path to a null child.

The term

**Rotations**

To balance a tree, you need to physically rearrange the nodes. If all the nodes are on the left of the root, for example, you need to move some of them over to the right side. This is done using

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• Raise some nodes and lower others to help balance the tree.

• Ensure that the characteristics of a binary search tree are not violated.

Recall that in a binary search tree the left children of any node have key values less than the node, while its right children have key values greater than or equal to the node. If the rotation didn’t maintain a valid binary search tree, it wouldn’t be of much use, because the search algorithm, as we saw in the preceding chapter, relies on the search-tree arrangement.

Note that color rules and node color changes are used only to help decide when to perform a rotation. Fiddling with the colors doesn’t accomplish anything by itself; it’s the rotation that’s the heavy hitter. Color rules are like rules of thumb for building a house (such as “exterior doors open inward”), while rotations are like the hammering and sawing needed to actually build it.

**Simple Rotations**

In Experiment 2 we tried rotations to the left and right. Those rotations were easy to visualize because they involved only three nodes. Let’s clarify some aspects of this process.

**What’s Rotating? **

The term

Remember that the top node isn’t the “center” of the rotation. If we talk about a car tire, the top node doesn’t correspond to the axle or the hubcap; it’s more like the topmost part of the tire tread.

The rotation we described in Experiment 2 was performed with the root as the top node, but of course any node can be the top node in a rotation, provided it has the appropriate child.

**Mind the Children**

You must be sure that, if you’re doing a right rotation, the top node has a left child.

Otherwise, there’s nothing to rotate into the top spot. Similarly, if you’re doing a left rotation, the top node must have a right child.

**The Weird Crossover Node**

Rotations can be more complicated than the three-node example we’ve discussed so far. Click Start, and then, with 50 already at the root, insert nodes with the following values, in this order: 25, 75, 12, 37.

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When you try to insert the 12, you’ll see the Can’t insert: needs color flip message. Just click the Flip button. The parent and children change color. Then press Ins again to complete the insertion of the 12. Finally, insert the 37. The resulting arrangement is shown in Figure 9.9a.

50

a)

25

75

Crossover node

12

37

25

b)

12

50

Crossover node

37

75

**FIGURE 9.9**

Rotation with crossover node.

Now we’ll try a rotation. Place the arrow on the root (don’t forget this!) and press the RoR button. All the nodes move. The 12 follows the 25 up, and the 50 follows the 75 down.

But what’s this? The 37 has detached itself from the 25, whose right child it was, and become instead the left child of 50. Some nodes go up, some nodes go down, but the 37 moves

In the original position of Figure 9.9a, the 37 is called an

It’s like becoming your own uncle (although it’s best not to dwell too long on this analogy).

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**Subtrees on the Move**

We’ve shown individual nodes changing position during a rotation, but entire subtrees can move as well. To see this, click Start to put 50 at the root, and then insert the following sequence of nodes in order: 25, 75, 12, 37, 62, 87, 6, 18, 31, 43.

Click Flip whenever you can’t complete an insertion because of the Can’t insert: needs color flip message. The resulting arrangement is shown in Figure 9.10a.

50

a)

25

75

12

37

62

87

25

6

18

31

43

25

b)

50

12

37

75

6

18

25

31

43

62

87

**FIGURE 9.10**

Subtree motion during rotation.

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Position the arrow on the root, 50. Now press RoR. Wow! (Or is it WoW?) A lot of nodes have changed position. The result is shown in Figure 9.10b. Here’s what happens:

• The top node (50) goes to its right child.

• The top node’s left child (25) goes to the top.

• The entire subtree of which 12 is the root moves up.

• The entire subtree of which 37 is the root moves across to become the left child of 50.

• The entire subtree of which 75 is the root moves down.

You’ll see the Error: root must be black message, but you can ignore it for the moment. You can flip back and forth by alternately pressing RoR and RoL with the arrow on the top node. Do this and watch what happens to the subtrees, especially the one with 37 as its root.

In Figure 9.10, the subtrees are encircled by dotted triangles. Note that the relations of the nodes within each subtree are unaffected by the rotation. The entire subtree moves as a unit. The subtrees can be larger (have more descendants) than the three nodes we show in this example. No matter how many nodes there are in a subtree, they will all move together during a rotation.

**Human Beings Versus Computers**

At this point, you’ve learned pretty much all you need to know about what a rotation does. To cause a rotation, you position the arrow on the top node and then press RoR or RoL. Of course, in a real red-black tree insertion algorithm, rotations happen under program control, without human intervention.

Notice however that, in your capacity as a human being, you could probably balance any tree just by looking at it and performing appropriate rotations. Whenever a node has a lot of left descendants and not too many right ones, you rotate it right, and vice versa.

Unfortunately, computers aren’t very good at “just looking” at a pattern. They work better if they can follow a few simple rules. T