本文共 1395 字，大约阅读时间需要 4 分钟。

题目链接:

Time limit: 3.000 seconds

Description

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

Problem solving report:

Description: 给你由'[',']','(',')'组成的字符串，问能否进行括号匹配。

Problem solving: 利用栈的性质，只要遇到能匹配的就移出栈，否则就进栈，最后再判断一下栈是否为空就行了。

#include 
   
    #include 
    
     #include 
     
      using namespace std;int main(){    int t;    char str[130];    scanf("%d%*c", &t);    while (t--)    {        gets(str);        stack 
      
        s;        for (int i = 0; str[i]; i++)        {            if (!s.empty())            {                if ((s.top() != '[' || str[i] != ']') && (s.top() != '(' || str[i] != ')'))                    s.push(str[i]);                else s.pop();            }            else s.push(str[i]);        }        if (!s.empty())            printf("No\n");        else printf("Yes\n");    }    return 0;}
      
     
    
   

 

转载地址：https://lzyws739307453.blog.csdn.net/article/details/86665075 如侵犯您的版权，请留言回复原文章的地址，我们会给您删除此文章，给您带来不便请您谅解！