题解 | #最长的括号子串#(细节状态转移&动态规划)

最长括号子串分为两种形式..()()()和...((...))

dp[i] :代表以i结尾的最长括号子串

• ...（）dp[i]=dp[i-2]+2 当是s[i-1]='('
• ...((..)) dp[i]=dp[i-1]+dp[i-dp[i-1]-2]+2 当s[i-dp[i-1]-1]='('

class Solution {
public:
    int longestValidParentheses(string s) {
        int maxans = 0, n = s.length();
        vector<int> dp(n, 0);
        for (int i = 1; i < n; i++) {
            if (s[i] == ')') {
                if (s[i - 1] == '(') {
                    dp[i] = (i >= 2 ? dp[i - 2] : 0) + 2; // ...()
                } else if (i - dp[i - 1] > 0 && s[i - dp[i - 1] - 1] == '(') {
                    dp[i] = dp[i - 1] + ((i - dp[i - 1]) >= 2 ? dp[i - dp[i - 1] - 2] : 0) + 2;
                }
                maxans = max(maxans, dp[i]);
            }
        }
        return maxans;
    }
};