<Sicily>Brackets Matching
一、题目描述
Let us define a regular brackets sequence in the following way:
Empty sequence is a regular sequence.
If S is a regular sequence, then (S) , [S] and {S} are both regular sequences.
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;
}