leetcode32

 class Solution {
 public:
     int longestValidParentheses(string s) {
         int n = s.length(), longest = ;
         stack<int> st;
         for (int i = ; i < n; i++) {
             if (s[i] == '(') st.push(i);
             else {
                 if (!st.empty()) {
                     if (s[st.top()] == '(') st.pop();
                     else st.push(i);
                 }
                 else st.push(i);
             }
         }
         if (st.empty()) longest = n;
         else {
             int a = n, b = ;
             while (!st.empty()) {
                 b = st.top(); st.pop();
                 longest = max(longest, a-b-);
                 a = b;
             }
             longest = max(longest, a);
         }
         return longest;
     }
 };

参考：https://leetcode.com/problems/longest-valid-parentheses/discuss/14126/My-O(n)-solution-using-a-stack

补充一个python的实现：

 class Solution:
     def longestValidParentheses(self, s: str) -> int:
         n = len(s)
         longest = 0
         st = []
         for i in range(n):
             if s[i] == '(':
                 st.append(i)
             else:#s[i] == ')'
                 if len(st) > 0 and s[st[-1]] == '(':#前一位置是'('
                     st.pop(-1)#获得一个合法匹配
                 else:
                     st.append(i)
         if len(st) == 0:#s所有字符都是合法括号对
             longest = n
         else:
             a,b = n,0
             while len(st) != 0:#st中记录的都是无法匹配的'('和')'出现的位置，可称为'单身括号'
                 b = st.pop(-1)
                 longest = max(longest,a-b-1)#计算当前'单身括号'与下一个'单身括号'的距离，就是最长合法substring(连续子串)的长度
                 a = b
             longest = max(longest,a)
         return longest