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Valid Parentheses 

Given a string containing just the characters ‘(‘, ‘)‘, ‘{‘, ‘}‘, ‘[‘ and ‘]‘, determine if the input string is valid.

The brackets must close in the correct order, "()" and "()[]{}" are all valid but "(]" and "([)]" are not.

关于堆栈的知识回顾：

#include <stack>

empty() 堆栈为空则返回真

pop() 移除栈顶元素

push() 在栈顶增加元素

size() 返回栈中元素数目

top() 返回栈顶元素

stack<int> a   定义一个堆栈

//vs2012测试代码
#include<iostream>
#include<string>
#include<stack>

using namespace std;

class Solution {
public:
    bool isValid(string s)
	{
		stack<char> temp;
		for(int i=0; i<s.length(); i++)
		{
			if( !temp.empty() )
			{
				if( temp.top()=='(' && s[i]==')'
					|| temp.top()=='[' && s[i]==']'
					|| temp.top()=='{' && s[i]=='}')
						temp.pop();
				else
					temp.push(s[i]);
			}
			else
				temp.push(s[i]);
		}
		return temp.empty();
    }
};

int main()
{
	string s;
	getline( cin , s);  //或者cin>>s;
	Solution lin;
	cout<<lin.isValid(s)<<endl;

	return 0;
}

//方法一：自测Accepted
class Solution {
public:
    bool isValid(string s)
	{
		stack<char> temp;
		for(int i=0; i<s.length(); i++)
		{
			if( !temp.empty() )
			{
				if( temp.top()=='(' && s[i]==')'
					|| temp.top()=='[' && s[i]==']'
					|| temp.top()=='{' && s[i]=='}')
						temp.pop();
				else
					temp.push(s[i]);
			}
			else
				temp.push(s[i]);
		}
		return temp.empty();
    }
};

//方法二：其他人版本
class Solution {
public:
    bool matchBracket(char a, char b)
    {
        if(a == '[' && b == ']') return true;
        if(a == '(' && b == ')') return true;
        if(a == '{' && b == '}') return true;
        return false;
    }
    bool isValid(string s) {
        stack<char> charStack;
        for(int i = 0; i < s.size(); ++i)
        {
            if(charStack.empty() || !matchBracket(charStack.top(), s[i])) charStack.push(s[i]);
            else charStack.pop();
        }

        return charStack.empty();
    }
};