NYOJ2括号配对问题

括号配对是最基本的栈的问题，它是栈入门的经典题目，思路是，如果是左括号直接进栈，如果是右括号，这时就要比较栈顶的元素与他是否匹配，如果匹配则出栈，否则进栈，下面是代码的实现：

 1 #include <stdio.h>
 2 #include <stdlib.h>
 3 typedef struct stack{//定义栈来存储括号
 4     char ch;
 5     struct stack *next;
 6 }link_stack;
 7 link_stack * init_link_stack();
 8 link_stack * push_stack(link_stack *top, char ch);
 9 link_stack * pop_stack(link_stack *top);
10 int main()
11 {
12     int n;
13     scanf("%d", &n);
14     getchar();
15     while(n --)
16     {
17         link_stack * top;
18         top = init_link_stack();//初始化top指针
19         char ch[10001];
20         scanf("%s", ch); int i = 0;
21         while(ch[i] != ‘\0‘)//判断读到结束符
22         {
23             if(((ch[i] == ‘]‘) && (top -> ch == ‘[‘))||((ch[i] == ‘)‘) &&(top -> ch == ‘(‘)))//如果将要进栈的是右括号，判断栈顶元素是否为左括号，如果是就弹出
24                 top = pop_stack(top);
25             else
26                 top = push_stack(top, ch[i]);//否则压栈
27             i ++;
28         }
29         if(top -> ch == ‘0‘)//判断栈是否为空
30             printf("Yes\n");
31         else
32             printf("No\n");
33     }
34     return 0;
35 }
36
37 link_stack * init_link_stack()//初始化栈函数
38 {
39     link_stack *node;
40     node = (link_stack *) malloc(sizeof(link_stack));
41     node -> next = NULL;
42     node ->ch = ‘0‘;
43     return node;
44 }
45 link_stack * push_stack(link_stack *top, char ch)//入栈函数
46 {
47     link_stack *node;
48     node = init_link_stack();
49     node -> ch = ch;
50     node -> next = top;
51     top = node;
52     return top;
53 }
54 link_stack * pop_stack(link_stack *top)//出栈函数
55 {
56     link_stack *node;
57     if(top -> next == NULL)
58         return top;
59     else
60     {
61         node = top;
62         top = top -> next;
63         free(node);
64         return top;
65
66     }
67
68 }