2-5 Two Stacks In One Array (20分)（在一个数组中实现两个堆栈）

Write routines to implement two stacks using only one array. Your stack routines should not declare an overflow unless every slot in the array is used.

Format of functions:

Stack CreateStack( int MaxElements ); int IsEmpty( Stack S, int Stacknum ); int IsFull( Stack S ); int Push( ElementType X, Stack S, int Stacknum ); ElementType Top_Pop( Stack S, int Stacknum );

where int Stacknum is the index of a stack which is either 1 or 2; int MaxElements is the size of the stack array; and Stack is defined as the following:

typedef struct StackRecord Stack; struct StackRecord { int Capacity; / maximum size of the stack array / int Top1; / top pointer for Stack 1 / int Top2; / top pointer for Stack 2 */ ElementType Array; / space for the two stacks */ }

Note: Push is supposed to return 1 if the operation can be done successfully, or 0 if fails. If the stack is empty, Top_Pop must return ERROR which is defined by the judge program.

Sample program of judge:

#include <stdio.h> #include <stdlib.h> #define ERROR 1e8 typedef int ElementType; typedef enum { push, pop, end } Operation; typedef struct StackRecord *Stack; struct StackRecord { int Capacity; /* maximum size of the stack array */ int Top1; /* top pointer for Stack 1 */ int Top2; /* top pointer for Stack 2 */ ElementType *Array; /* space for the two stacks */ }; Stack CreateStack( int MaxElements ); int IsEmpty( Stack S, int Stacknum ); int IsFull( Stack S ); int Push( ElementType X, Stack S, int Stacknum ); ElementType Top_Pop( Stack S, int Stacknum ); Operation GetOp(); /* details omitted */ void PrintStack( Stack S, int Stacknum ); /* details omitted */ int main() { int N, Sn, X; Stack S; int done = 0; scanf("%d", &N); S = CreateStack(N); while ( !done ) { switch( GetOp() ) { case push: scanf("%d %d", &Sn, &X); if (!Push(X, S, Sn)) printf("Stack %d is Full!\n", Sn); break; case pop: scanf("%d", &Sn); X = Top_Pop(S, Sn); if ( X==ERROR ) printf("Stack %d is Empty!\n", Sn); break; case end: PrintStack(S, 1); PrintStack(S, 2); done = 1; break; } } return 0; } /* Your function will be put here */

Sample Input:

5 Push 1 1 Pop 2 Push 2 11 Push 1 2 Push 2 12 Pop 1 Push 2 13 Push 2 14 Push 1 3 Pop 2 End

Sample Output:

Stack 2 is Empty! Stack 1 is Full! Pop from Stack 1: 1 Pop from Stack 2: 13 12 11

为了充分利用顺序存储形式的堆栈的空间，采用在一个数组中实现两个堆栈的方式，哪个没满就继续放。下面附上这道题的中文链接，是之前做的

6-7 在一个数组中实现两个堆栈 (20分)（附思路分析和易错点分析）

Stack CreateStack(int MaxElements){ Stack s = (Stack)malloc(sizeof(struct StackRecord)); s->Array = (ElementType *)malloc(sizeof(ElementType)*MaxElements); s->Capacity = MaxElements; s->Top1 = -1; s->Top2 = MaxElements; return s; } int IsEmpty(Stack S,int Stacknum){ if(Stacknum==1&&S->Top1==-1) return 1; if(Stacknum==2&&S->Top2==S->Capacity) return 1; return 0; } int IsFull(Stack S){ if(S->Top2-S->Top1==1) return 1; return 0; } int Push(ElementType X, Stack S,int Stacknum){ if(IsFull(S)) return 0; if(Stacknum==1) S->Array[++(S->Top1)] = X; else S->Array[--(S->Top2)] = X; return 1; } ElementType Top_Pop(Stack S,int Stacknum){ if(IsEmpty(S,Stacknum)) return ERROR; if(Stacknum==1) return S->Array[(S->Top1)--]; else return S->Array[(S->Top2)++]; }