stacks chapter 8. objectives in this chapter, you will: learn about stacks examine various stack...
TRANSCRIPT
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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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
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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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top()
T& top(){ if (arrayList<T>::empty()) throw stackEmpty(); return get(arrayList<T>::size() - 1);}
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push(theElement)
void push(const T& theElement){insert(arrayList<T>::size(), theElement);}
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pop()
void pop(){ if (arrayList<T>::empty()) throw stackEmpty(); erase(arrayList<T>::size() - 1);}
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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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top
pop()
void pop(){ if (stackTop == -1) throw stackEmpty(); stack[stackTop--].~T(); // destructor for T}
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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 C
1234
• 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/Brahma
A B C
123
Towers Of Hanoi/Brahma
• 3-disk Towers Of Hanoi/Brahma
A B C
12
3
Towers Of Hanoi/Brahma
• 3-disk Towers Of Hanoi/Brahma
A B C
1 2 3
Towers Of Hanoi/Brahma
• 3-disk Towers Of Hanoi/Brahma
A B C
1 23
Towers Of Hanoi/Brahma
• 3-disk Towers Of Hanoi/Brahma
A B C
123
Towers Of Hanoi/Brahma
• 3-disk Towers Of Hanoi/Brahma
A B C
123
Towers Of Hanoi/Brahma
• 3-disk Towers Of Hanoi/Brahma
A B C
12
3
Towers Of Hanoi/Brahma
• 3-disk Towers Of Hanoi/Brahma
A B C
123
• 7 disk moves
Recursive Solution
A B C
1
• 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 C
1
• move top disk from A to C
Recursive Solution
A B C
1
• move top n-1 disks from B to C using A
Recursive Solution
A B C
1
• 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