Stacks

Chapter 8

3

Objectives

In this chapter, you will:• Learn about stacks• Examine various stack operations• Learn how to implement a stack as an array• Learn how to implement a stack as a linked list• Discover stack applications• Learn how to use a stack to remove recursion

C++ Programming: Program Design Including Data Structures, Fifth

Edition

Stacks• Stack: list of homogenous elements

– Addition and deletion occur only at one end, called the top of the stack

– Example: in a cafeteria, the second tray can be removed only if first tray has been removed

– Last in first out (LIFO) data structure • Operations:

– Push: to add an element onto the stack – Pop: to remove an element from the stack

Stacks

• Linear list.• One end is called top.• Other end is called bottom.• Additions to and removals from the top end only.

top

bottom

Various types of stacks

Stack Of Cups

• Add a cup to the stack.

bottom

top

C

A

B

D

E

F

• Remove a cup from new stack.• A stack is a LIFO list.

bottom

top

C

A

B

D

E

The Abstract Class stack

template<class T>class stack { public: virtual ~stack() {} virtual bool empty() const = 0; virtual int size() const = 0; virtual T& top() = 0; virtual void pop() = 0; virtual void push(const T& theElement) = 0;};

Derive From A Linear List Class

• Class arrayList• Class chain

– stack top is either left end or right end of linear list– empty() => arrayList::empty()– size() => arrayList::size()– top() => get(0) or get(size() - 1)

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a b c d e

Derive From arrayList

Derive From arrayList

• when top is left end of linear list–push(theElement) => insert(0, theElement)–pop() => erase(0)–use left end of list as top of stack

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a b c d e

Derive From arrayList

–when top is right end of linear list• push(theElement) => insert(size(), theElement)• pop() => erase(size()-1)

–use right end of list as top of stack

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a b c d e

Derive From Chain

–stack top is either left end or right end of linear list

–empty() => chain::empty()–size() => chain::size()

a b c d e

NULL

firstNode

Derive From Chain

a b c d e

NULL

firstNode

– when top is left end of linear list top() => get(0)push(theElement) => insert(0, theElement)pop() => erase(0)

Derive From Chain

a b c d enull

firstNode

– when top is right end of linear list• top() => get(size() - 1)• push(theElement) => insert(size(), theElement)• pop() => erase(size()-1)•use left end of list as top of stack

Derive From arrayList

template<class T>class derivedArrayStack : private arrayList<T>, public stack<T>{ public: // code for stack methods comes here };

Constructor

derivedArrayStack(int initialCapacity = 10) : arrayList<T> (initialCapacity) {}

empty() And size()

bool empty() const{return arrayList<T>::empty();}

int size() const{return arrayList<T>::size();}

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a b c d e

top()

T& top(){ if (arrayList<T>::empty()) throw stackEmpty(); return get(arrayList<T>::size() - 1);}

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a b c d e

push(theElement)

void push(const T& theElement){insert(arrayList<T>::size(), theElement);}

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a b c d e

pop()

void pop(){ if (arrayList<T>::empty()) throw stackEmpty(); erase(arrayList<T>::size() - 1);}

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a b c d e

template<class T>class derivedArrayStack : private arrayList<T>,public stack<T>{ public: derivedArrayStack(int initialCapacity = 10) :arrayList<T>(initialCapacity) {} bool empty() const {return arrayList<T>::empty();} int size() const {return arrayList<T>::size();} T& top() { if (arrayList<T>::empty()) throw stackEmpty(); return get(arrayList<T>::size() - 1); } void pop() { if (arrayList<T>::empty()) throw stackEmpty(); erase(arrayList<T>::size() - 1); } void push(const T& theElement) {insert(arrayList<T>::size(), theElement);}};

Evaluation• Merits of deriving from arrayList

– Code for derived class is quite simple and easy to develop.

– Code is expected to require little debugging.– Code for other stack implementations such as a

linked implementation are easily obtained.• Just replace private arrayList<T> with private chain<T>• For efficiency reasons we must also make changes to use

the left end of the list as the stack top rather than the right end.

Code From Scratch• Use a 1D array stack whose data type is T.

– same as using array element in arrayList• Use an int variable stackTop.

– Stack elements are in stack[0:stackTop].– Top element is in stack[stackTop].– Bottom element is in stack[0].– Stack is empty iff stackTop = -1.– Number of elements in stack is stackTop + 1.

Code From Scratchtemplate class<T>class arrayStack : public stack<T>{ public: // public methods come here private: int stackTop; // current top of stack int arrayLength; // stack capacity T *stack; // element array};

Constructor

template<class T>arrayStack<T>::arrayStack(int initialCapacity){// Constructor. if (initialCapacity < 1) {// code to throw an exception comes here } arrayLength = initialCapacity; stack = new T[arrayLength]; stackTop = -1;}

push(…)

template<class T>void arrayStack<T>::push(const T& theElement){// Add theElement to stack. if (stackTop == arrayLength - 1) {// code to double capacity coms here } // add at stack top stack[++stackTop] = theElement;}

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a b c d e

top

pop()

void pop(){ if (stackTop == -1) throw stackEmpty(); stack[stackTop--].~T(); // destructor for T}

0 1 2 3 4

a b c d e

top

Linked Stacktemplate<class T>class linkedStack : public stack<T>{ public: linkedStack(int initialCapacity = 10) {stackTop = NULL; stackSize = 0;} ~linkedStack(); bool empty() const {return stackSize == 0;} int size() const {return stackSize;} T& top() { if (stackSize == 0) throw stackEmpty();

return stackTop->element; } void pop(); void push(const T& theElement);private: chainNode<T>* stackTop; // pointer to stack top int stackSize; // number of elements in stack};

Push()

template<class T>void linkedStack<T>::push(const T& theElement) { stackTop = new chainNode<T>(theElement, stackTop); stackSize++; }

Pop()template<class T>void linkedStack<T>::pop(){// Delete top element. if (stackSize == 0) throw stackEmpty();

chainNode<T>* nextNode = stackTop->next; delete stackTop; stackTop = nextNode; stackSize--;}

Implement stack

• derivedArrayStack.h• derivedLinkedStack.h• arrayStack.h• linkedStack.h

Stack in C++

Applications

Parentheses Matching• (((a+b)*c+d-e)/(f+g)-(h+j)*(k-l))/(m-n)

– Output pairs (u,v) such that the left parenthesis at position u is matched with the right parenthesis at v.• (2,6) (1,13) (15,19) (21,25) (27,31) (0,32) (34,38)

• (a+b))*((c+d)– (0,4)– right parenthesis at 5 has no matching left parenthesis– (8,12)– left parenthesis at 7 has no matching right parenthesis

Parentheses Matching• scan expression from left to right• when a left parenthesis is encountered, add its

position to the stack• when a right parenthesis is encountered, remove

matching position from stack

Example• (((a+b)*c+d-e)/(f+g)-(h+j)*(k-l))/(m-n)

0 1 2

Example• (((a+b)*c+d-e)/(f+g)-(h+j)*(k-l))/(m-n)

0 1

(2,6)

Example• (((a+b)*c+d-e)/(f+g)-(h+j)*(k-l))/(m-n)

0 1

(2,6) (1,13)15

Example• (((a+b)*c+d-e)/(f+g)-(h+j)*(k-l))/(m-n)

0 1

(2,6) (1,13) (15,19)21

Example• (((a+b)*c+d-e)/(f+g)-(h+j)*(k-l))/(m-n)

0 1

(2,6) (1,13) (15,19) (21,25)27

Example• (((a+b)*c+d-e)/(f+g)-(h+j)*(k-l))/(m-n)

0 1

(2,6) (1,13) (15,19) (21,25)(27,31) (0,32)

• and so on

Homework 2.1

• Write a program that look for matched pairs of parentheses and matched pairs of brackets ([]) and report a nesting problem.

• For example: in the string (a+[b*(c-d)+f]), the output should be (0,14),(3,13),(6,10).

Towers Of Hanoi/Brahma

A B C1234

• 64 gold disks to be moved from tower A to tower C• each tower operates as a stack• cannot place big disk on top of a smaller one

Towers Of Hanoi/Brahma

• 3-disk Towers Of Hanoi/BrahmaA B C123

Towers Of Hanoi/Brahma

• 3-disk Towers Of Hanoi/BrahmaA B C12

3

Towers Of Hanoi/Brahma

• 3-disk Towers Of Hanoi/BrahmaA B C1 2 3

Towers Of Hanoi/Brahma

• 3-disk Towers Of Hanoi/BrahmaA B C1 2

3

Towers Of Hanoi/Brahma

• 3-disk Towers Of Hanoi/BrahmaA B C

123

Towers Of Hanoi/Brahma

• 3-disk Towers Of Hanoi/BrahmaA B C

123

Towers Of Hanoi/Brahma

• 3-disk Towers Of Hanoi/BrahmaA B C

12

3

Towers Of Hanoi/Brahma

• 3-disk Towers Of Hanoi/BrahmaA B C

123

• 7 disk moves

Recursive Solution

A B C1

• n > 0 gold disks to be moved from A to C using B• move top n-1 disks from A to B using C

Recursive Solution

A B C1

• move top disk from A to C

Recursive Solution

A B C1

• move top n-1 disks from B to C using A

Recursive Solution

A B C1

• moves(n) = 0 when n = 0• moves(n) = 2*moves(n-1) + 1 = 2n-1 when n > 0

Homework 2.2• 2.2.1: Write a recursive version of program

to solve n-disk Tower of Hanoi problem• 2.2.2: determine the maximum number of

disk you can solve using recursive and stack programs.

Rat In A Maze

Rat In A Maze

• Move order is: right, down, left, up• Block positions to avoid revisit.

Rat In A Maze

• Move order is: right, down, left, up• Block positions to avoid revisit.

Rat In A Maze

• Move backward until we reach a square from which a forward move is possible.

Rat In A Maze

• Move down.

Rat In A Maze

• Move left.

Rat In A Maze

• Move down.

Rat In A Maze

• Move backward until we reach a square from which a forward move is possible.

Rat In A Maze

• Move backward until we reach a square from which a forward move is possible.

• Move downward.

Rat In A Maze

• Move right.• Backtrack.

Rat In A Maze

• Move downward.

Rat In A Maze

• Move right.

Rat In A Maze

• Move one down and then right.

Rat In A Maze

• Move one up and then right.

Rat In A Maze

• Move down to exit and eat cheese.• Path from maze entry to current position operates as

a stack.

Homework 2.3 (bonus)

• Implement the Rat in a maze

Homework 2• Due date: Feb 24