class Solution {public:    int longestValidParentheses(string s) {        vector
     
       stack;        int maxlen = 0;        int curlen = 0;        int last = -1;        int len = s.length();        for (int i=0; i
      
        maxlen) maxlen = curlen;        }                return maxlen;    }};
      
     

参考：http://www.cnblogs.com/zhuli19901106/p/3547419.html

last用来记录最后一个无法匹配的右括号的位置，当stack为空时，必然存在一个合法的括号序列，而这个序列的起始肯定是last后面的一个位置。

智商有限，做了好久，只有看答案了。。。

 

第二轮：

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

For "(()", the longest valid parentheses substring is "()", which has length = 2.

Another example is ")()())", where the longest valid parentheses substring is "()()", which has length = 4.

还是没想起来，真是硬伤啊，真是硬伤。。。

class Solution {public:    int longestValidParentheses(string s) {        int len = s.size();        if (len < 2) return 0;        int last_invalid = -1;                int pos = 0;        int maxlen = 0;        vector
     
       idx;        while (pos