6-7 在一个数组中实现两个堆栈

6-7 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
思路：该指针是一维的，所以要用数组实现两个堆栈。我有两种想法申请一个2*MaxSize+1的数组，从中间开始分别加减实现两个栈，但是碰到输出超限，于是我打算补齐两个 /* details omitted */函数（细节决定成败，你却给我省略了）\脑疼🤒；然后试了第二种申请大小为MaxSize的数组，从两头开始向中。然而还是输出超限。。。。我要去百度了拜拜~
看完答案之后—>这种题目目的是为了让我们猜输出函数？？？简直浪费表情
两细节省略函数如下：

Operation GetOp()
{
    char Push[] = "Push";
    char Pop[] = "Pop";
    char End[] = "End";
    char s[100];
    scanf("%s", s);

    if (strcmp(Push, s) == 0)return push;
    if (strcmp(Pop, s) == 0)return pop;
    if (strcmp(End, s) == 0)return end;
}
void PrintStack(Stack S, int Tag)
{
    printf("Pop from Stack %d:", Tag);
    if (Tag == 1){
        while (S->Top1 != -1){
            printf(" %d", S->Data[S->Top1--]);
        }
    }
    else {
        while (S->Top2 != S->MaxSize){
            printf(" %d", S->Data[S->Top2++]);
        }
    }
    putchar('\n');
}

AC代码如下：

 

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

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