4-7 在一个数组中实现两个堆栈 (20分)

4-7 在一个数组中实现两个堆栈 (20分)

本题要求在一个数组中实现两个堆栈。

函数接口定义：

Stack CreateStack( int MaxSize );
bool Push( Stack S, ElementType X, int Tag );
ElementType Pop( Stack S, int Tag );

其中 Tag 是堆栈编号，取1或2； MaxSize 堆栈数组的规模； Stack 结构定义如下：

typedef int Position;
struct SNode {
ElementType *Data;
Position Top1, Top2;
int MaxSize;
};
typedef struct SNode *Stack;

注意：如果堆栈已满， Push 函数必须输出“Stack Full”并且返回false；如果某堆栈是空的，则 Pop 函数必须输出“Stack Tag Empty”（其中Tag是该堆栈的编号），并且返回ERROR。

裁判测试程序样例：

#include <stdio.h>
#include <stdlib.h>

#define ERROR 1e8
typedef int ElementType;
typedef enum { push, pop, end } Operation;
typedef enum { false, true } bool;
typedef int Position;
struct SNode {
ElementType *Data;
Position Top1, Top2;
int MaxSize;
};
typedef struct SNode *Stack;

Stack CreateStack( int MaxSize );
bool Push( Stack S, ElementType X, int Tag );
ElementType Pop( Stack S, int Tag );

Operation GetOp();  /* details omitted */
void PrintStack( Stack S, int Tag ); /* details omitted */

int main()
{
int N, Tag, X;
Stack S;
int done = 0;

scanf("%d", &N);
S = CreateStack(N);
while ( !done ) {
switch( GetOp() ) {
case push:
scanf("%d %d", &Tag, &X);
if (!Push(S, X, Tag)) printf("Stack %d is Full!\n", Tag);
break;
case pop:
scanf("%d", &Tag);
X = Pop(S, Tag);
if ( X==ERROR ) printf("Stack %d is Empty!\n", Tag);
break;
case end:
PrintStack(S, 1);
PrintStack(S, 2);
done = 1;
break;
}
}
return 0;
}

/* 你的代码将被嵌在这里 */

输入样例：

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

输出样例：

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

点击访问 PAT-测验

Stack CreateStack( int MaxSize )
{
Stack S = (Stack)malloc(sizeof(struct SNode));
S->MaxSize = MaxSize;
S->Data = (ElementType *)malloc(sizeof(ElementType)*MaxSize);
S->Top2 = MaxSize;
S->Top1 = -1;
return S;
}
bool Push(Stack S, ElementType X, int Tag){
if(S->Top1+1 == S->Top2){
printf("Stack Full\n");
return false;
}else{
if(Tag == 1){
S->Top1++;
S->Data[S->Top1] = X;
}else{
S->Top2--;
S->Data[S->Top2] = X;
}
}
return true;
}

ElementType Pop(Stack S, int Tag){
int X;
if(Tag == 1){
if(S->Top1 == -1){
printf("Stack 1 Empty\n");
return ERROR;
}else{
X = S->Data[S->Top1];
S->Top1--;
return X;
}
}else if(Tag == 2){
if(S->Top2 == S->MaxSize){
printf("Stack 2 Empty\n");
return ERROR;
}else{
X = S->Data[S->Top2];
S->Top2++;
return X;
}
}
}