/*
    棧 鏈式儲存結構實現
    第一個結構體是每一個節點元素, 類似於連結串列 
    第二個結構體是用個指標top指向連結串列的尾部 
*/
#include<stdio.h>
#include<stdlib.h>
#define OK 1
#define ERROR 0

typedef int SElemType;
typedef int Status;

typedef struct StackNode{
    SElemType data; 
    struct StackNode *next; 
} StackNode, *LinkStackPtr;

typedef
 
 struct LinkStack{
    LinkStackPtr top;
    int count;
} LinkStack;

//初始化棧
Status initStack(LinkStack *S){ 
    LinkStackPtr p;
    p = (LinkStackPtr)malloc(sizeof(StackNode));
    p->next = NULL;
    S->top = p;
    S->count = 0;
    return OK;
} 

//判斷棧是否為空
Status StackEmpty(LinkStack S){
    if
 
(S.count == 0){
        return OK;  //為null 
    }else{
        return ERROR;   
    }
}
//入棧 插入元素e
Status push(LinkStack *S, SElemType e){
    LinkStackPtr s = (LinkStackPtr) malloc (sizeof(StackNode));     //生成一個節點
    s->data = e;
    s->next = S->top;   //把當前的棧頂元素值賦給新節點的直接後繼
    S->top = s;     //將新的節點s賦值給棧頂指標
 

    S->count++;
    return OK;  
} 

//出棧,刪除棧頂元素 用e返回出棧的值
Status pop(LinkStack *S, SElemType *e){
    //判斷棧是否為空 
    if(StackEmpty(*S)){
        return ERROR;
    }
    *e = S->top->data;  //儲存要出棧的元素 
    LinkStackPtr p;
    p = S->top;     //將棧頂結點賦值給p 
    S->top = S->top->next;  //棧頂指向下一個結點 
    free(p);    //釋放結點p
    S->count--;
    return OK; 
} 

//遍歷棧
void stackTraverse(LinkStack S) {
    if(S.count == 0){
        printf("棧為空\n");
    }else{
        LinkStackPtr p;
        p = S.top;
        while(p->next){
            printf("%d ",p->data);
            p = p->next;
        }
        printf("\n");
    }
}

int main(void){
    LinkStack S;
    SElemType e; 
    initStack(&S);
    printf("\n 1.初始化棧 \n 2.遍歷棧 \n 3.入棧 \n 4.出棧 \n 0.退出\n");
    int option = 1, value;
    while(option){
        scanf("%d",&option);
        switch(option){
            case 1:
                initStack(&S);
                break;
            case 2:
                stackTraverse(S);
                break;
            case 3:
                printf("請輸入要入棧的資料\n");
                scanf("%d",&e);
                push(&S, e); 
                stackTraverse(S);
                break;
            case 4:
                value = pop(&S, &e);
                if(value == 0){
                    printf("沒有可用的元素可以出棧\n");
                }else{
                    printf("出棧的元素是:%d\n",e);
                } 
                break;
            case 0:
                return OK;
        }
    }
    return OK;
}