Leetcode: 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.

匹配之类的应该要想到堆栈，再用一个DP来记录以i结尾的有效字符串的最大长度。其实自己弄复杂了，堆栈里面存放index的话会简单一些。

class Solution {
public:
    int longestValidParentheses(string s) {
        int size = s.size();
        int *dp = new int[size];
        int max_length = 0;
        stack<char> parens;
        for (int i = 0; i < size; ++i) {
            dp[i] = 0;
            if (s[i] == '(') {
                parens.push(s[i]);
            }
            else {
                if (!parens.empty()) {
                    char tmp = parens.top();
                    parens.pop();
                    dp[i] = 2;
                    if (dp[i-1] > 0) {
                        dp[i] += dp[i-1];
                    }
                    int prev = i - dp[i];
                    if (prev >= 0 && dp[prev] > 0) {
                        dp[i] += dp[prev];
                    }
                    
                    if (dp[i] > max_length) {
                        max_length = dp[i];
                    }
                }
            }
        }
        
        delete []dp;
        return max_length;
    }
};

堆栈里面存放索引，确实简单些。

class Solution {
public:
    int longestValidParentheses(string s) {
        int result = 0;
        vector<int> lens(s.size(), 0);
        stack<int> indices;
        for (int i = 0; i < s.size(); ++i) {
            if (s[i] == '(') {
                indices.push(i);
            }
            else {
                if (!indices.empty()) {
                    int len = i - indices.top() + 1;
                    if (i - len >= 0) {
                        len += lens[i-len];
                    }
                    lens[i] = len;
                    if (len > result) {
                        result = len;
                    }
                    indices.pop();
                }
            }
        }
        
        return result;
    }
};