673 - Parentheses Balance

最新推荐文章于 2019-02-07 09:12:08 发布

杜庆彦

最新推荐文章于 2019-02-07 09:12:08 发布

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分类专栏： uva ACM

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本文介绍了一个简单的程序，用于检查由括号组成的字符串是否正确配对。该程序通过使用栈来跟踪括号，确保每对括号都能正确闭合。

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Parentheses Balance

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

#include 
#include 
#define N 129
int judge(char s[])
{
    char stack[N];//可以不用此数组，可直接在数组s中操作。
    int top = 0;
    int i, f;
    f = 1;
    for(i = 0; s[i] != '\0'; i++){
        if(s[i] == '[' || s[i] == '(')
            stack[top++] = s[i];
        else if(top > 0 && ((s[i] == ']' && stack[top-1] == '[') ||
                  (s[i] == ')' && stack[top-1] == '(')))
                top--;
        else{
            f = 0; break;
        }
    }
    if(top != 0)
        f = 0;
    return f;
}
int main()
{
    int n;
    char s[N];
    scanf("%d", &n);
    gets(s);      
    while(n-- > 0){
        gets(s);
        if(judge(s))
            printf("Yes\n");
        else
            printf("No\n");
    }
    return 0;
}