leetcode-20 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.

简单利用stack。
时间为0ms.
注意：当栈为空时，是不能检测栈顶的，因为此时根本没有栈顶，调试显示为进入非法区域。这点，容易忽略，切记切记！

/*
想法是，遇到左括号，全部进栈，遇到右括号，检查栈顶是否匹配，匹配，去掉栈顶，不匹配，进栈，直到string末尾，看栈是否为空，为空，合法，不为空，不合法。
*/

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