6-7 在一个数组中实现两个堆栈
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
总结:
1、两个堆栈指针相向而行
2、数组的动态分配:(数组的大小为m)
eg: int* Array = (int*)malloc( m * sizeof(int) );
eg:
struct SNode {
    ElementType *Data;
    Position Top1, Top2;
    int MaxSize;
};
typedef struct SNode *Stack;
//动态分配
Stack stack=(Stack)malloc(sizeof(struct SNode));
stack->Data=(int* )malloc((MaxSize)*sizeof(int));
3、引入C++标准库 using namespace std;
字符串类型string(s小写)属于c++标准库,C语言中是没有的。
string类型变量 使用cin cout 输出输出, 或者使用以下方法printf输出
string word = "hanjiale";
printf("%s", word.c_str());
原本自己写的代码:(报错:输出超限)
我觉得没有什么问题,可能是PrintStack()出错了。
/* 你的代码将被嵌在这里 */ Stack CreateStack( int MaxSize ){ Stack stack=(Stack)malloc(sizeof(struct SNode)); stack->MaxSize=MaxSize; stack->Data=(int* )malloc((MaxSize+1)*sizeof(int)); stack->Top1=0; stack->Top2=MaxSize-1; return stack; } bool Push( Stack S, ElementType X, int Tag ){ if (!S) return false; if(Tag==1){ if(S->Top1>S->Top2){ printf("Stack Full\n"); return false; } S->Data[S->Top1]=X; S->Top1++; return true; }else { if(S->Top2<S->Top1){ printf("Stack Full\n"); return false; } S->Data[S->Top2]=X; S->Top2--; return true; } } ElementType Pop( Stack S, int Tag ){ if (!S) return false; if(Tag==1){ if(S->Top1==0){ printf("Stack 1 Empty\n"); return ERROR; } return S->Data[S->Top1--]; }else { if(S->Top2==S->MaxSize-1){ printf("Stack 2 Empty\n"); return ERROR; } return S->Data[S->Top2++]; } }
正确代码:
Stack CreateStack( int MaxSize ){ Stack stack=(Stack)malloc(sizeof(struct SNode)); stack->MaxSize=MaxSize; stack->Data=(int* )malloc((MaxSize+1)*sizeof(int)); stack->Top1=-1; stack->Top2=MaxSize; return stack; } bool Push( Stack S, ElementType X, int Tag ){ if (!S) return false; if(Tag==1){ if(S->Top1+1==S->Top2){ printf("Stack Full\n"); return false; } S->Top1++; S->Data[S->Top1]=X; return true; }else { if(S->Top2-1==S->Top1){ printf("Stack Full\n"); return false; } S->Top2--; S->Data[S->Top2]=X; return true; } } ElementType Pop( Stack S, int Tag ){ if (!S) return false; if(Tag==1){ if(S->Top1==-1){ printf("Stack 1 Empty\n"); return ERROR; } return S->Data[S->Top1--]; }else { if(S->Top2==S->MaxSize){ printf("Stack 2 Empty\n"); return ERROR; } return S->Data[S->Top2++]; } }
 
                    
                     
                    
                 
                    
                
 
                
            
         
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浙公网安备 33010602011771号