02-線性結構4 Pop Sequence

　　這道題是2013年PAT春季考試真題，考察隊堆棧的基本概念的掌握。在保證輸入正確後，關鍵在於對"Pop"序列的判斷，我用isPopOrder函數進行了判斷，代碼如下：

#include <stdio.h>
#include <stdlib.h>
#define MaxSize 1001

typedef struct SNode *Stack;
struct SNode{
    int Data[MaxSize];
    int MaxCap;  //Maximum capacity of this stack
    int Top;
};

Stack CreateStack(int M)
{
    Stack PtrS;
    PtrS = (Stack)malloc(sizeof(struct SNode));
    PtrS->Top = -1;
    PtrS->MaxCap = M;
    return PtrS;
}

/*堆棧基本操作此處略去*/

int isPopOrder(int *CheckOrder, int M, int N, int K)
{
    int i, ci = 0;
    Stack S;
    S = CreateStack(M);
    for(i = 1; i < N + 1; i++){
        Push(i, S);  //Push in the order of 1, 2, ..., N
        while(CheckOrder[ci] == GetTop(S)){
            Pop(S);
            ci++;
        }
    }
    if(GetTop(S) == -1)  //堆棧為空
        return 1;
    else
        return 0;
}

int main(int argc, char const *argv[])
{
    int M, N, K;  //M is the maximum capacity of the stack; Push N numbers; K sequences to be checked
    int i, j;
    int CheckOrder[MaxSize], res[MaxSize];
    scanf("%d %d %d", &M, &N, &K);
    for(i = 0; i < K; i++){
        for(j = 0; j < N; j++){
            scanf("%d", &CheckOrder[j]);
        }
        res[i] = isPopOrder(CheckOrder, M, N, K);
    }
    for(i = 0; i < K; i++){
        if(res[i])
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}

（待更新）