Parentheses Balance

本文介绍了一个用于检查字符串中括号()和[]是否正确配对的C++程序。该程序通过栈来实现括号的匹配验证,并能处理多个字符串输入,判断每个字符串中的括号是否正确闭合。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

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. Yourprogram 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 stringa line.

Output

A sequence of ‘Yes’ or ‘No’ on the output file.

Sample Input

3

([])

(([()])))

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

Sample Output

Yes

No

Yes


#include<iostream>
#include<stack>
#include<cstdio>
using namespace std;
int pd(char a,char b)
{
	if(a=='('&&b==')')
		return 0;
	if(a=='['&&b==']')
		return 0;
	return 1;
}
int tj(char m)
{
	if(m=='['||m=='(')
		return 1;
	return 0;
}
int main()
{
	char s[200];
	int n;
	scanf("%d",&n);
	getchar();
	while(n--)
	{
		stack<char> a;
		gets(s);
		if(s[0]=='\n')
		{
			printf("Yes\n");
			continue;
		}
		for(int i=0;s[i];i++)
		{
			if(a.empty())
				a.push(s[i]);
			else if(pd(a.top(),s[i]))
			{
				if(tj(s[i]))
					a.push(s[i]);
			}
			else
				a.pop();
		}
		if(a.empty())
			printf("Yes\n");
		else
			printf("No\n");
	}
	return 0;
}

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值