在一个数组中实现两个堆栈

3-2(*) 在一个数组中实现两个堆栈 (10 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
结尾无空行

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
结尾无空行
/* 你的代码将被嵌在这里 */

输入样例:

4
Del
Add 5
Add 4
Add 3
Del
Del
Add 2
Add 1
Add 0
Add 10
End
结尾无空行

输出样例:

Queue Empty
5 is out
4 is out
Queue Full
3 2 1 0
结尾无空行

原代码(有错误):

Stack CreateStack( int MaxSize )
{
    Stack s=(Stack)malloc(sizeof(struct SNode));
    s->Top1=-1;
    s->Top2=MaxSize;
    s->Data=(ElementType*)malloc(sizeof(ElementType)*MaxSize);
    s->MaxSize=MaxSize;
    return s;
}
bool Push( Stack S, ElementType X, int Tag )
{
    if(Tag==1){
        if(S->Top2-S->Top1==1){
            printf("Stack Full\n");
            return false;
        }
       
        //*(S->Data)=X;
        S->Top1++;
        S->Data[S->Top1]=X;
        
        return true;
        
    }
    else{
        if(S->Top2==0){
            printf("Stack Full\n");
            return false;
        }
        (S->Top2)--;
        
        //*((S->Data)+(S->Top2)-(S->Top1)-1)=X;
        S->Data[S->Top2]=X;
    }
    return true;
}

ElementType Pop( Stack S, int Tag ){
     if(Tag==1){
        if(S->Top1==-1){
            printf("Stack %d Empty\n",Tag);
            return ERROR;
        }
        /* int x=S->Data[S->Top1];
         S->Data[S->Top1]=0;
         S->Top1--;
         return x;*/
         return S->Data[S->Top1++];
         
             
         
    }
    else{
        if((S->Top2)==(S->MaxSize)-1){
             printf("Stack %d Empty\n",Tag);
            return ERROR;
        }
        /*
        int x=S->Data[S->Top2];
        S->Data[S->Top2]=0;
        S->Top2++;
       return x;*/
        return S->Data[S->Top2++];
        //*((S->Data)+(S->Top2)-(S->Top1)-1)=0;
        
    }
}

代码里面两个致命的错误:1.栈满的标志应该是Top2-Top1=1;2栈2空的标志是Top2=MaxSize;

题目正式分析:
本题属于基于动态顺序存储结构的堆栈的实现,实际上我们在解决栈的空间溢出的问题时常常使两个栈共享空间,一个栈的栈尾在存储空间开头,另一个的栈尾在存储空间末尾。

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

ElementType Pop( Stack S, int Tag ){
     if(Tag==1){
        if(S->Top1==-1){
            printf("Stack %d Empty\n",Tag);
            return ERROR;
        }
     return S->Data[S->Top1--];
         
             
         
    }
    else{
        if((S->Top2)==(S->MaxSize)){
             printf("Stack %d Empty\n",Tag);
            return ERROR;
        }
 
       return S->Data[S->Top2++];
        
    }
}
上一篇:九度oj 1523 从上往下打印二叉树


下一篇:回溯法求解组合数独