Uva 673 Parentheses Balance 平衡的括号

You are given a string consisting of parentheses () and []. A string of this type is said to be correct:
(a) if it is the empty string
(b) if A and B are correct, AB is correct,
(c) if A is correct, (A ) and [A ] is correct.
Write a program that takes a sequence of strings of this type and check their correctness. Your program can assume that the maximum string length is 128.

Input

The file contains a positive integer n and a sequence of n strings of parentheses () and [], one string a line.

Output

A sequence of Yes or No on the output file.

Sample Input

3
( [ ] )
( ( [ ( ) ] ) ) )
( [ ( ) [ ] ( ) ] ) ( )

Sample Output

Yes
No
Yes

题意：输入包含小括号和中括号的括号序列，是否合法。
     空序列和括号匹配合法输出Yes 否则输出No；

实现代码

#include <cstdio>
#include <cstring>
const int  maxn = 130;
char str1[maxn];
char str2[maxn];

int main()
{
    int i, n, top, len;

    scanf("%d", &n);
    getchar();        //去除\n

    while (n--)
    {
        gets(str2);
        len = strlen(str2);
        if (len == 0)
        {
            printf("Yes\n");
            continue;
        }
        for (top = 0, i = 0; i<len; i++)
        {
            if (str2[i] != '(' && str2[i] != ')' && str2[i] != '[' && str2[i] != ']')   // 判断str2中的是否为括号符，否则退出此次循环
                continue;
            str1[top] = str2[i]; //将第一个括号符压入栈底，top指向栈底
            if (top == 0) // 栈底只有一个元素，使top指向第二个元素
            {
                top++;
                continue;
            }
            if ((str1[top - 1] == '(' && str1[top] == ')') || (str1[top - 1] == '[' &&str1[top] == ']')) //判断是否有匹配的括号符，则top指向栈前元素，否则top自加1
                top--;
            else
                top++;
        }
        if (top <= 0)
            printf("Yes\n"); //top=0或者-1 则符合平衡 输出Yes 否则输出No
        else
            printf("No\n");
    }
    return 0;
}