32. Longest Valid Parentheses

Longest Valid Parentheses

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.

解题思路：这道题有多种方法可以做，最后采用了动态规划。但普通的动态规划都是向前考虑的，这道题是向后考虑的，可以说是逆向的。比如普通动态规划dp[i]应该是从位置0~i的最长有效括号长度，而这道题的dp[i]表示的是从i到length-1的最长有效括号长度。
因此i从后往前推，若s[i]=’(‘，则匹配的右括号位置end应该是i + dp[i + 1] + 1，若该位置确实是右括号则dp[i] = dp[i + 1] + 2，另外为了计算连续的有效括号长度需要加上dp[end + 1]。

代码：

class Solution {
public:
    int longestValidParentheses(string s) {
        int len = s.length();
        int *dp = new int[len];
        memset(dp,0,len*sizeof(int));
        for (int i = len - 2; i >= 0; i--) {
            if (s[i] == '(') {
                int end = i + dp[i + 1] + 1;
                if (end<len && s[end] == ')') {
                    dp[i] = dp[i + 1] + 2;
                    if (end<len - 1) {
                        dp[i] += dp[end + 1];
                    }
                }
            }
        }
        int maxLen = 0;
        for (int i = 0; i<len; i++) {
            if (dp[i]>maxLen)
                maxLen = dp[i];
        }
        return sizeof(dp);
        delete[]dp; 
    }
};