Leetcode #32 Longest Valid Parentheses

最新推荐文章于 2024-02-09 15:39:44 发布

旻宇

最新推荐文章于 2024-02-09 15:39:44 发布

阅读量403

点赞数

CC 4.0 BY-SA版权

分类专栏：动态规划字符串文章标签： DP string

本文链接：https://blog.youkuaiyun.com/u010129448/article/details/47660111

动态规划同时被 2 个专栏收录

13 篇文章

订阅专栏

字符串

6 篇文章

订阅专栏

本文介绍了一种使用动态规划方法解决寻找字符串中最长的有效括号子串的问题。通过一个C++实现的例子，展示了如何遍历字符串并利用动态规划数组记录有效括号长度的过程，最终找到最长有效括号子串的长度。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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.

class Solution {
public:
    int longestValidParentheses(string s) {
        int n = s.size(), res = 0;
        vector<int> dp(n, 0);

        for (int i = 0; i < n; i++) {
            if (s[i] == ')' && i - 1 >= 0) {
                if (s[i - 1] == '(')
                    dp[i] = i >= 2 ? 2 + dp[i - 2] : 2;
                else if(i-1 - dp[i-1] >= 0 && s[i-1 - dp[i-1]] == '(')
                    dp[i] = dp[i-1] + 2 + (i-1 - dp[i -1] -1 >= 0 ? dp[i-1 - dp[i -1] -1] : 0);
                res = max(res, dp[i]);
            }
        }
        return res;
    }
};