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：

 #include <stdio.h>
 #include <stdlib.h>
 typedef struct stack{//定义栈来存储括号
     char ch;
     struct stack *next;
 }link_stack;
 link_stack * init_link_stack();
 link_stack * push_stack(link_stack *top, char ch);
 link_stack * pop_stack(link_stack *top);
 int main()
 {
     int n;
     scanf("%d", &n);
     getchar();
     while(n --)
     {
         link_stack * top;
         top = init_link_stack();//初始化top指针
         char ch[];
         scanf("%s", ch); int i = ;
         while(ch[i] != '\0')//判断读到结束符
         {
             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
                 top = pop_stack(top);
             else
                 top = push_stack(top, ch[i]);//否则压栈
             i ++;
         }
         if(top -> ch == '')//判断栈是否为空
             printf("Yes\n");
         else
             printf("No\n");
     }
     return ;
 }

 link_stack * init_link_stack()//初始化栈函数
 {
     link_stack *node;
     node = (link_stack *) malloc(sizeof(link_stack));
     node -> next = NULL;
     node ->ch = '';
     return node;
 }
 link_stack * push_stack(link_stack *top, char ch)//入栈函数
 {
     link_stack *node;
     node = init_link_stack();
     node -> ch = ch;
     node -> next = top;
     top = node;
     return top;
 }
 link_stack * pop_stack(link_stack *top)//出栈函数
 {
     link_stack *node;
     if(top -> next == NULL)
         return top;
     else
     {
         node = top;
         top = top -> next;
         free(node);
         return top;

     }

 }