一、题目描述

Let us define a regular brackets sequence in the following way:

1. Empty sequence is a regular sequence.

2. If S is a regular sequence, then (S) , [S] and {S} are both regular sequences.

3. If A and B are regular sequences, then AB is a regular sequence.

For example, all of the following sequences of characters are regular brackets sequences:

(), [], {}, (){[]}

While the following character sequences are not:

(, [, {, )(, ([)], {[(]

Write a program to judge whether a given sequence is a regular bracket sequence or not.

二、输入

The input may contain several test cases.

The first line of each test case contains the length of the bracket sequence N (1<=N<=100). The second line contains N characters including ‘(‘, ‘)’, ‘[‘, ‘]’,’{‘ or ’}’.

Input is terminated by EOF.

三、输出

For each test case, if the sequence is a regular brackets sequence, output “YES” on a single line, else output “NO”.

例如：
输入：
2
()
4
((]]
6
{[]}()
输出：
YES
NO
YES

四、解题思路

这个使用栈来处理，当栈顶不为空的时候，栈顶跟输入的字符（括号）匹配，是否为一对（栈顶的为左边括号，新输入的为右边括号）。若能匹配，pop出栈顶，继续新的输入，否则新输入的字符入栈。

五、代码

#include<iostream>
#include<stack>
#include <stdio.h>using namespace std;int main()
{int leng;while(cin >> leng){stack<char> strStack;for(int i = 0; i < leng; i++){char nowChar;cin >> nowChar;if(strStack.empty()){strStack.push(nowChar); continue;} //如果栈为空，直接入栈，不需要判断if(nowChar == '(' || nowChar  == '[' || nowChar == '{'){strStack.push(nowChar); continue;}  //如果是左边的括号，直接入栈if(nowChar == ')' && strStack.top() == '('){strStack.pop();}    //“()”匹配成功else if(nowChar == ']' && strStack.top() == '['){strStack.pop();}       //“[]”匹配成功else if(nowChar == '}' && strStack.top() == '{') {strStack.pop();}      //“{}”匹配成功else {strStack.push(nowChar);}  //匹配不成功，入栈}if(strStack.empty()) cout << "YES" << endl;else cout << "NO" << endl;}return 0;
}