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/ / Stack class - - array implementation // / / CONSTRUCTION: with no parameters // / / ******************puBLIC OPERATIONS*********************
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void push( x ) void pop ( ) Object top( Object topAndPop( boo1 isEmpty( ) void makeEmpty ( )
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x most recently inserted item most recently inserted item and remove most recent item true if empty; else false all items
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/ / UnderflowException thrown as needed
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template <class Object> class Stack
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public: Stack(
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boo1 isEmpty ( ) const; const Object & top( ) const; void makeEmpty( ) ; void pop ( ) ; void push ( const Object Object topAndPop ( ) ;
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private: vector<Object> theArray; int topofstack;
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Interface for the array-based Stack class
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/ / Construct the stack template <class Object> Stack<Object>::Stack( )
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The zero-parameter constructor for the array-based S t a c k class
Stacks and Queues
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/ / Test if the stack is logically empty / / Return true if empty, false, otherwise
template <class Object> boo1 Stack<Object>::isEmpty( ) const
return topofstack == -1; 1
/ / Make the stack logically empty template <class Object> void Stack<Object>::makeEmpty( )
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topofstack = -1; 1
The i s E m p t y and makeEmpty routines for the array-based S t a c k class
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/ / Insert x into the stack template <class Object> void Stack<Object>::push( const Object
if( topofstack == theArraysize( ) - 1 theArrayresize( theArraysize( ) * 2 + 1 theArray[ ++topofstack I = x;
push method for the array-based S t a c k class
can charge the O(N) cost of the doubling over these N / 2 easy pushes, thereby effectively raising the cost of each p u s h by only a small constant This technique is known as amortization A real-life example of amortization is payment of income taxes Rather than pay your entire bill on April 15, the government requires that you pay most of your taxes through withholding The total tax bill is always the same; it is when the tax is paid that varies The same is true for the time spent in the p u s h operations We can charge for the array doubling at the time it occurs, or we can bill each p u s h operation equally An amortized bound requires that we bill each operation in a sequence for its fair share of the total cost In our example, the cost of array doubling therefore is not excessive
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Dynamic Array Implementations
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/ / Return the most recently inserted item in the stack
/ / Does not alter the stack / / Throws UnderflowException if stack is already empty
template <class Object> const Object & Stack<Object>: :top(
const
if ( isEmpty( j ) throw UnderflowException( return theArray[ topofstack 1 ;
/ / Remove the most recently inserted item from the stack / / Throw UnderflowException if stack is already empty
template <class Object> void Stack<Object>::pop()
if ( isEmpty( ) ) throw UnderflowException( j ; topofstack--;
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/ / Return and remove most recently inserted item from the stack / / Throws UnderflowException if stack is already empty
template <class Object> Object Stack<Object>: topAndPop ( ) :
if ( isEmpty ( ) j throw UnderflowException( ) ; return theArray[ topofstack-- I ;
The topAndPop method for the array-based Stack class
1612 Queues
The easiest way to implement the queue is to store the items in an array with the front item in the front position (ie, array index 0) If back represents the position of the last item in the queue, then to enqueue we merely increment back and place the item there The problem is that the d e q u e u e operation is very expensive The reason is that, by requiring that the items be placed at the start of the array, we force the d e q u e u e to shift all the items one position after we remove the front item