LeetCode 题解（111）: Longest Valid Parentheses

题目：

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.

题解：

首先想到的是动态规划，但是这道题的递推公式不太好理解。而用一个stack比较好理解。

C++版：

class Solution {public:    int longestValidParentheses(string s) {        int n = s.size(), maxLen = 0;        vector<int> dp(n+1,0);        for(int i=1; i<=n; i++) {            int j = i-2-dp[i-1];            if(s[i-1]=='(' || j<0 || s[j]==')')                 dp[i] = 0;            else {                dp[i] = dp[i-1]+2+dp[j];                maxLen = max(maxLen, dp[i]);            }        }        return maxLen;    }};

Java版：

public class Solution {    public int longestValidParentheses(String s) {        Stack<Integer> left = new Stack<>();        int overall = 0;        for(int i = 0; i < s.length(); i++) {            if(s.charAt(i) == '(') {                left.push(i);            } else {                if(!left.empty() && s.charAt(left.peek()) == '(') {                    left.pop();                    overall = Math.max(overall, (left.empty() ? i + 1 : i - left.peek()));                } else {                    left.push(i);                }            }        }        return overall;    }}

Python版：

class Solution:    # @param {string} s    # @return {integer}    def longestValidParentheses(self, s):        overall = 0        l = []        for i in range(len(s)):            if s[i] == '(':                l.append(i)            else:                if len(l) != 0 and s[l[-1]] == '(':                    l = l[:-1]                    if len(l) == 0:                        overall = max(overall, i + 1)                    else:                        overall = max(overall, i - l[-1])                else:                    l.append(i)        return overall