Problem Solving with Data Structures using Java: A Multimedia Approach

Problem Solving with Data Structures using Java: A Multimedia Approach

Chapter 11: Abstract Data Types: Separating the

Meaning from the Implementation

Chapter Objectives

Separating Meaning from Implementation

Powerful engineering idea in computer science.• Meaning

• Identify the behavior of an object or data structure.

• Build other program components to interact with that behavior.

• Implementation• Hide how that behavior is implemented, so that no

program components can be dependent on the implementation.

Abstract Data Type: Stacks

An example of an Abstract Data Type (ADT):• We can define the methods and the

behavior apart from any implementation.

• There are multiple implementations, some better than others.

Can use a stack to reverse a list in less time, but more space.• A classic tradeoff!

Space for time.

Only put on top,Never pull from

bottom.

A Stack is a LIFO List

Last-In-First-Out (LIFO) List• First item in the list is the last one out.

• Last one in is first one out.

I got here first!

I got here second!

I got here third!

This is the top of the stack

New items go at the top

I got here first!

I got here second!

I got here third!

This is the new top of the stack

I got here fourth!

Items only get removed from the top

I got here first!

I got here second!

I got here third!

This is the new (er, old) top of the stack

I got here fourth! And now I’m outta here!

What can we do with stacks?

push(anObject): Tack a new object onto the top of the stack

pop(): Pull the top (head) object off the stack.

peek(): Get the top of the stack, but don’t remove it from the stack.

size(): Return the size of the stack

Separation of Concerns

Separation definition of behavior of data structure from implementation.

Lets one change without changing the other.

Java allows us to use an interface as a way specify that behavior definition.

Stack Interface

/**

* An abstract definition of a stack

* @author Barb Ericson

*/

public interface Stack<E> {

/**

* Method to add the element to the top of the stack

* @param element the element to add

*/

public void push(E element);

/**

* Method to return the top element on the

* stack, but not remove it from the stack

* @return the top object from the stack

*/

public E peek();

Just defining the operations here.

E is the type of the things in the Stack.

Stack<E> represents the generic stack for any time element

Just defining the operations here.

E is the type of the things in the Stack.

Stack<E> represents the generic stack for any time element

/**

* Method to remove the top element from the

* stack and return it

* @return the top element from the stack

*/

public E pop();

/**

* Method to return the number of elements in

* the stack

* @return the number of elements in the stack

*/

public int size();

/**

* Method to check if the stack is empty

* @return true if empty, else false

*/

public boolean isEmpty();

}

Creating an implementation

We declare a class as implementing that interface.

We can continue to use the <E> generic, so that we can later create a stack for any type object.

Stack as a LinkedList

import java.util.LinkedList; // Need for LinkedList

/**

* Class that represents a stack using a linked list

* of objects

* @author Mark Guzdial


*/

public class LinkedListStack<E> implements Stack<E> {

/** Where we store the elements */

private LinkedList<E> elements;

/**

* Constructor that takes no arguments

*/

public LinkedListStack() {

elements = new LinkedList<E>();

}

//// Methods ///

/**

* Method to add an element to the stack


*/

public void push(E element) {

// New elements go at the front

elements.addFirst(element);

}

/**

* Method to return the top element on the stack

* but leave the element on the stack

* @return the top element on the stack

*/

public E peek() {

return elements.getFirst();

}

/**

* Method to remove the top element from a stack

* and return it

* @return the top element from the stack and remove it

*/

public E pop() {

E toReturn = this.peek();

elements.removeFirst();

return toReturn;

}

/**

* Method to get the number of elements in the stack


*/

public int size(){return elements.size();}

/**

* Method to test if the stack is empty

* @return true if the stack is empty, else false

*/

public boolean isEmpty() {

return (size() == 0);

}

}

Using the Stack with Strings > Stack<String> stack = new LinkedListStack<String>();

> stack.push("This")

> stack.push("is")

> stack.push("a")

> stack.push("test")

> stack.size()

4

> stack.peek()

"test"

> stack.pop()

"test"

> stack.pop()

"a"

> stack.pop()

"is"

> stack.pop()

"This"

Notice the use of <String>

Notice the use of <String>

A Stack ofPictures

> Stack<Picture> stack = new LinkedListStack<Picture>();

> stack.push(new Picture(FileChooser.getMediaPath("beach.jpg")));

> stack.push(new Picture(FileChooser.getMediaPath("arch.jpg")));

> stack.push(new Picture(FileChooser.getMediaPath("bridge.jpg")));

> stack.size()

3

> stack.peek()

Picture, filename C:\dsBook\media-source/bridge.jpg

height 640 width 480

> stack.pop()

Picture, filename C:\dsBook\media-source/bridge.jpg


> stack.pop()

Picture, filename C:\dsBook\media-source/arch.jpg


> stack.pop()

Picture, filename C:\dsBook\media-source/beach.jpg


Note: You can’t create an object using an interface name, e.g., new Stack<String>()

Note: You can’t create an object using an interface name, e.g., new Stack<String>()

A stack is a stack, no matter what lies beneath. Our description of the stack minus the

implementation is an example of an abstract data type (ADT).• An abstract type is a description of the methods that a

data structure knows and what the methods do. We can actually write programs that use the

abstract data type without specifying the implementation. • There are actually many implementations that will

work for the given ADT.• Some are better than others.

New implementation: Stack as Array

/**

* Implementation of a stack as an array



*/

public class ArrayStack<E> implements Stack<E> {

/** default size of the array */

private static final int ARRAY_SIZE = 20;

/** Where we'll store our elements */

private Object[] elements;

Constructor

/** Index where the top of the stack is */

private int top;

/**

* No argument constructor

*/

public ArrayStack() {

elements = new Object[ARRAY_SIZE];

top = 0;

}

Methods for Stack as Array

/**

* Method to add an element to the top of the stack


*/


// New elements go at the top

elements[top]=element;

// then add to the top

top++;

if (top==ARRAY_SIZE) {

System.out.println("Stack overflow!");

}

}

/**


* but not remove it.

* @return the object at the top of the stack

*/

public E peek() {

if (top==0) {

System.out.println("Stack empty!");

return null;

} else {

// this will give a warning but it is unavoidable

return (E) elements[top-1];

}

}

/**

* Method to remove and return the top element on the stack

* @return the element on the top of the stack

*/

public E pop() {


top--;

return toReturn;

}

/**

* Method to return the number of elements in the stack


*/

public int size(){return top;}

/**

* Method to check if the stack is empty

* @return true if the stack is empty else false

*/

public boolean isEmpty() {return this.size() == 0;}

}

Trying out Stack as Array

> Stack<String> stack = new ArrayStack<String>();

Pushing one element

> stack.push("Matt");

Push two more, pop one

> stack.push("Katie");

> stack.push("Jenny");

> stack.size() // without the ending ';' it prints out the result

3

> stack.peek() // without the ending ';' it prints out the result

"Jenny"

> stack.pop()

"Jenny"

Critique the array-based implementation

What happens if the number of elements in the stack is more than 20?

What are stacks good for?

The algorithm for converting an equation into a tree uses a stack.

Often use stacks when describing card games, like solitaire.

The list of Web pages you have visited in a browser is stored in a stack, so that you can go “back.”

The list of operations you have executed is stored in a stack, so that you can “undo.”

Stacks describe function calls

As new functions get called, position in old functions get pushed on a stack.• So you always return to the last function you

were in.

• If an error occurs, you get a stack trace.

• If your recursion goes into an infinite loop, what error do you get? Stack overflow!

A Stack Example: New Reverse

Recall our original implementation of reverse().• We go to the end of the original list to find the last().

• We then remove() it (which involves walking the list until we find the one before last())

• We then insert it at the end of the new list (via add(), which does last().insertAfter()).

All told: For each node, we walk the whole list three times.• O(n*n2)=O(n3)

Original Reverse

/** * Reverse the list starting at this, * and return the last element of the list. * The last element becomes the FIRST element * of the list, and THIS points to null. */ public LayeredSceneElement reverse() { LayeredSceneElement reversed, temp;

// Handle the first node outside the loop reversed = this.last(); this.remove(reversed);

while (this.getNext() != null) { temp = this.last(); this.remove(temp); reversed.add(temp); }

// Now put the head of the old list on the end of // the reversed list. reversed.add(this);

// At this point, reversed // is the head of the list return reversed; }

Highly inefficient.Touching each node requires touching every other node: O(n2)

Highly inefficient.Touching each node requires touching every other node: O(n2)

New Reverse: Push all, pull off reversed

/**

* Reverse2: Push all the elements on

* the stack, then pop all the elements

* off the stack.

*/

public LayeredSceneElement reverse2() {

LayeredSceneElement reversed, current, popped;

Stack<LayeredSceneElement> stack =

new LinkedListStack<LayeredSceneElement>();

// Push all the elements on the list

current=this;

while (current != null)

{

stack.push(current);

current = current.getNext();

}

// Make the last element (current top of stack) into new first

reversed = stack.pop();

// Now, pop them all onto the list

current = reversed;

while (stack.size()>0) {

popped = stack.pop();

current.insertAfter(popped);

current = popped;

}

return reversed;

}

What’s the diff? Time

How often is each node touched in reverse2()?• Twice: Once going onto the stack, once

coming off.

• O(2*n) => O(n)

The stack-based reverse is faster than the original reverse.

What’s the diff? Space

How much space does reverse2() take?• Whatever space the stack takes.

How much additional space does reverse() take? None

Very common tradeoff: Space for time.• You can make an algorithm go faster, by using more

space.

• If you need to fit into less memory, you have to do more processing, which takes more time.

Testing Reverse in SoundListTest() public void reverseTest(){ Sound s = null; // For copying in sounds s = new Sound(FileChooser.getMediaPath("guzdial.wav")); SoundNode root = new SoundNode(s); s = new Sound(FileChooser.getMediaPath("is.wav")); SoundNode one = new SoundNode(s); root.last().insertAfter(one); s = new Sound(FileChooser.getMediaPath("scritch-q.wav")); SoundNode two = new SoundNode(s); root.last().insertAfter(two); s = new Sound(FileChooser.getMediaPath("clap-q.wav")); SoundNode three = new SoundNode(s); two.insertAfter(three); //root.playFromMeOn(); SoundNode reversed = (SoundNode) root.reverse2(); reversed.playFromMeOn(); }

Second ADT: Introducing a Queue

First-In-First-Out List• First person in line is first person served

I got here first!

I got here second!

I got here third!

This is the front or head of the queue

This is the tail of the queue

First-in-First-out

New items only get added to the tail.• Never in the middle

Items only get removed from the head.

I got here first!

I got here second!

I got here third!

This is the front or head of the queue


As items leave, the head shifts

I got here first! AND NOW I’M UP!

I got here second!

I got here third!

Now, this is the front or head of the queue


Served!

As new items come in, the tail shifts

I got here second!

I got here third!

Now, this is the front or head of the queue

Now, this is the tail of the queue

I got here fourth!

Queue Operations

push(element): Tack a new element onto the tail (end) of the queue.

pop(): Pull the top (head) element off the queue.

peek(): Get the head of the queue, but don’t remove it from the queue.

size(): Return the size of the queue. isEmpty(): Return true or false, if the size of

the queue is zero.

Queue Interface

/** * Interface to define an abstract queue * @author Barb Ericson */public interface Queue<E> {

/** * Push an element onto the tail of the Queue * @param element the element to add to the queue */ public void push(E element);

/** * Peek at, but don't remove, the head of the queue * @return the head of the queue (top) */ public E peek();

/**

* Pop an object from the Queue

* @return the head (top) of the queue and

* remove it from the queue

*/

public E pop();

/**

* Return the size of a queue

* @return the number of elements in the queue

*/

public int size();

/**

* Method to see if the queue is empty

* @return true if the queue is empty, else false

*/

public boolean isEmpty();

}

Implementing a Queue as a Linked List

import java.util.*; // LinkedList representation

/**

* Implements a simple queue using a linked list



*/

public class LinkedListQueue<E> extends AbstractQueue<E> {


private LinkedList<E> elements;

/**


*/

public LinkedListQueue() {

elements = new LinkedList<E>();

}

/// Methods

/**

* Push an element onto the tail of the Queue

* @param element the element to add to the queue

*/


elements.addFirst(element);

}

/**

* Peek at, but don't remove, top (first) of queue

* @return the first object in the queue

*/

public E peek() {

return elements.getLast();

}

//**


* @return the top object from the queue (and remove it)

*/

public E pop() {


elements.removeLast();

return toReturn;

}

/**



*/

public int size() { return elements.size(); }

/**



*/

public boolean isEmpty() { return size() == 0; }

}

Testing our implementation:Behaving as expected?

> Queue<String> line = new LinkedListQueue<String>();

> line.push("Fred");

> line.push("Mary");

> line.push("Jose");

> line.size()

3

> line.peek() // without ending ';' prints the result

"Fred"

> line.pop()

"Fred"

> line.peek()

"Mary"

> line.pop()

"Mary"

> line.peek()

"Jose"

> line.pop()

"Jose"

Queue as Array

/**

* Implements a simple queue using an array



*/

public class ArrayQueue<E> extends AbstractQueue<E> {

/** constant for the size of the queue */

private static final int ARRAY_SIZE = 20;


private Object[] elements;

/** The index of the head */

private int head;

/** The index of the tail */

private int tail;

/** No argument constructor */

public ArrayQueue() {

elements = new Object[ARRAY_SIZE];

head = 0; tail = 0;

}

/// Methods

/**

* Push an element onto the tail of the Queue


*/


if ((tail + 1) >= ARRAY_SIZE) {

System.out.println("Queue underlying implementation failed");}

else {

// Store at the tail,

// then increment to a new open position

elements[tail] = element;

tail++;}

}

/**

* Peek at, but don't remove, the head of the queue

* @return the head of the queue (top)

*/

public E peek() {

// this will give a warning but there is no way around it

return (E) elements[head];

}

/**




*/

public E pop() {


if (((head + 1) >= ARRAY_SIZE) ||

(head > tail)) {

System.out.println("Queue underlying implementation failed.");

return toReturn;

}

else {

// Increment the head forward, too.

head++;

return toReturn;

}

}

/**



*/

public int size() { return tail-head;}

/**



*/

public boolean isEmpty() { return size() == 0; }

}

Again, testing implementation

> Queue<String> line = new ArrayQueue<String>();

> line.push("Fred");

> line.push("Mary");

> line.push("Jose");

> line.size()

3

> line.peek() // without ending ';' prints the result

"Fred"

> line.pop()

"Fred"

> line.peek()

"Mary"

> line.pop()

"Mary"

> line.peek()

"Jose"

> line.pop()

"Jose"

How the array implementation of queue works

An empty queue

Pushing “Matt”

Pushing “Katie”

Popping (returns “Matt”)

Notice that we’ve now permanently “lost” the first cell.Challenge: Can you “recover” that cell?Notice that we’ve now permanently “lost” the first cell.Challenge: Can you “recover” that cell?

Improving the implementation

Our two implementations have duplicated code.• Check out isEmpty() in each.

Where could we put the code so that it’s not duplicated?• Interfaces can’t have method bodies.

• We can use an abstract class.

Abstract Queue Class

/**

* Class to define an abstract queue


*/

public abstract class AbstractQueue<E> implements Queue<E> {

/**

* Push an object onto the Queue


*/

public abstract void push(E element);

/**

* Peek at, but don't remove, the head of the queue

* @return the head of the queue (top)

*/

public abstract E peek();

/**




*/

public abstract E pop();

/**



*/

public abstract int size();

Finally, the isEmpty method, factored out

/**



*/

public boolean isEmpty() {

return (size() == 0);

}

}

Revised ArrayQueue

/**

* Implements a simple queue using an array



*/

public class ArrayQueue extends AbstractQueue {

/// ... fields and other methods as before

/**



*/

// commented out since inherited from AbstractQueue

// public boolean isEmpty() { return size() == 0; }

}

Revised LinkedListQueue

/**

* Implements a simple queue using a linked list



*/

public class LinkedListQueue extends AbstractQueue {

// ... fields and other methods as before

/**

* Check if the queue is empty

* @return true if no elements in the queue, else false

*/

// commented out since inherited from AbstractQueue

// public boolean isEmpty() {return this.size() == 0;}

}

Can switch implementations easily

As ArrayList

> import java.util.*;

> List<String> nameList = new ArrayList<String>();

> nameList.add("Shayna");

> nameList.add("Marcus");

> nameList.add("Jakita");

> nameList

[Shayna, Marcus, Jakita]

If inserting/deleting a lot, use LinkedList

> import java.util.*;

> List<String> nameList = new LinkedList<String>();

> nameList.add("Shayna");

> nameList.add("Marcus");

> nameList.add("Jakita");

> nameList

[Shayna, Marcus, Jakita]

How the List is used (methods and behavior) is the same in both cases.

How the List is used (methods and behavior) is the same in both cases.

A Common Pattern

We see this structure repeated throughout java.util• Interface defines ADT

• Abstract class defines common parts

• Concrete classes differ in implementation

Switching implementations

Switching the underlying implementation can provide advantages.• Of speed

• Of flexibility

For example, our stack implemented as an Array can go out of bounds if grows too large.• But not if we use ArrayList!

Using an ArrayList:ArrayListStack

import java.util.*;

/**

* Implementation of a stack as an ArrayList



*/

public class ArrayListStack<E> implements Stack<E> {


private List<E> list = new ArrayList<E>();

/**


*/

public ArrayListStack() {

}

//// Methods ///

/**

* Method to add an element to the top of the stack


*/


list.add(element);

}

/**


* but not remove it.

* @return the object at the top of the stack

*/

public E peek() {

return list.get(list.size() - 1);

}

/**

* Method to remove and return the top element on the stack

* @return the element on the top of the stack

*/

public E pop() {

return list.remove(list.size() - 1);

}

/**

* Method to return the number of elements in the stack


*/

public int size(){return list.size();}

Main() for testing

public static void main(String[] args) {

Stack<String> stack = new ArrayListStack<String>();

stack.push("Matt");

stack.push("Katie");

stack.push("Jenny");

System.out.println(stack.size());

System.out.println(stack.peek());

System.out.println(stack.pop());



}

}

New ADT: A Map

A Map associates a value with a key.• The key can be any unique object.

• The value can be any single object.

If you store two values under one key, the second value will overwrite the first value as THE value for THAT key.

Using a Map to create oil paintings

Oil paintings have only one color for region (size of brush)

Using the Java implementation of Map (java.util.Map) to track how often a color (the key) appears in a region.

For each region,• Store color counts in map.

• Get most common (highest count) color.

• Set all colors in region to the most common color.

oilPaint method in Picture

/**

* Method to do an oil paint effect on a picture

* @param dist the distance from the current pixel

* to use in the range

* @return the new picture

*/

public Picture oilPaint(int dist) {

// create the picture to return

Picture retPict = new Picture(this.getWidth(),this.getHeight());

// declare pixels

Pixel currPixel = null;

Pixel retPixel = null;

// loop through the pixels

for (int x = 0; x < this.getWidth(); x++) {

for (int y = 0; y < this.getHeight(); y++) {

currPixel = this.getPixel(x,y);

retPixel = retPict.getPixel(x,y);

retPixel.setColor(currPixel.getMostCommonColorInRange(dist));

}

}

return retPict;

}

getMostCommonColorInRange in class Pixel

/**

* Method to return the most common color in the given

* range from the current pixel in the picture or just

* return this color.

* @param dist the distance to use for the range

* @return the most common color in this range

*/

public Color getMostCommonColorInRange(int dist) {

Map<Color,Integer> colorMap = new HashMap<Color,Integer>();

Pixel currPixel = null;

Integer value = null;

Color theKey = null;

The variable colorMap is of type Map.We instantiate HashMap, but could switch to another implementation

The variable colorMap is of type Map.We instantiate HashMap, but could switch to another implementation

// loop through the pixels around this one within the distance

for (int currY = y - dist; currY <= y + dist; currY++) {

for (int currX = x - dist; currX <= x + dist; currX++) {

if (currY >= 0 && currY < picture.getHeight() &&

currX >= 0 && currX < picture.getWidth()) {

currPixel = picture.getPixel(currX,currY);

theKey = currPixel.getColor();

value = colorMap.get(theKey);

if (value == null)

colorMap.put(theKey,1);

else

colorMap.put(theKey, value + 1);

}

}

}

// find the color that is most common

int maxValue = 1;

int currValue = 0;

theKey = this.getColor(); // use current color as default

Set<Color> keySet = colorMap.keySet();

for (Color key : keySet) {

currValue = colorMap.get(key);

if (currValue > maxValue) {

theKey = key;

maxValue = currValue;

}

}

return theKey;

}

Example Use

Documents

Problem Solving with Data Structures using Java: A Multimedia Approach