題目：

Given a string containing just the characters '(' and ')', find the length of the longest valid (well-formed) parentheses substring.

Example 1:

Input: "(()"
Output: 2
Explanation: The longest valid parentheses substring is "()"

Example 2:

Input: ")()())"
Output: 4
Explanation:
 
 The longest valid parentheses substring is "()()"

程式碼：

class Solution {
public:
    int longestValidParentheses(string s) {
     int sum = 0, record = 0, count = 0, max1 = INT_MIN,max2=INT_MIN;
	int len = s.length();
	bool flag = false;

	for (int i = 0; i < len; i++) {
		if (s[i] == '(') {
			sum++;
			record++;
			flag = true;
		}
		if (flag&&s[i] == ')') {
			sum--;
			if (sum == 0) {
				count += record * 2;
				record = 0;
			}
			if (sum < 0) {
				max1 = max1 > count ? max1 : count;
				count = 0;
                sum=0;
				flag = false;
			}
		}
	}
	max1 = max1 > count ? max1 : count;
	sum = 0; record = 0; count = 0; flag = false;

	for (int i = len - 1; i >= 0; i--) {
		if (s[i] == ')') {
			sum++;
			record++;
			flag = true;
		}
		if (flag&&s[i] == '(') {
			sum--;
			if (sum == 0) { count += record * 2; record = 0; }
			if (sum < 0) {
				max2 = max2 > count ? max2 : count;
				count = 0;
                sum=0;
				flag = false;
			}
		}
	}

	max2 = max2 > count ? max2 : count;
	return max1 > max2 ? max1 : max2;
    }
};
 

思路：

1、用一個數字給左括號，右括號計數

     往右走的時候左括號加右括號減，往左走記右括號加左括號減。
     計數小於零的時候重置計數
     一路下來記錄最長的就完事了。

2、從向左方向和向右方向都需要遍歷

3、值得注意的點：當計數小於0的時候要給它重置為0