Given a string s, find the longest palindromic substring in s. You may assume that the maximum length of s is 1000. Example 1: Input: "babad" Output: "bab" Note: "aba" is also a valid answer. Example 2: Input: "cbbd"…

问题描述: 给定一个字符串S,找出它的最大的回文子串,你可以假设字符串的最大长度是1000,而且存在唯一的最长回文子串 . 思路分析: 动态规划的思路:dp[i][j] 表示的是 从i 到 j 的字串,是否是回文串. 则根据回文的规则我们可以知道: 如果s[i] == s[j] 那么是否是回文决定于 dp[i+1][ j - 1] 当 s[i] != s[j] 的时候, dp[i][j] 直接就是 false. 动态规划的进行是按照字符串的长度从1 到 n推进的. DP算法实现: package…

Longest Palindromic Substring -- HARD 级别 Question SolutionGiven a string S, find the longest palindromic substring in S. You may assume that the maximum length of S is 1000, and there exists one unique longest palindromic substring. 经典的DP题目. 主页君给出3种解…

Given a string S, find the longest palindromic substring in S. You may assume that the maximum length of S is 1000, and there exists one unique longest palindromic substring. 思路I:动态规划,遍历到i的时候要保证之前的元素都已经计算过状态,所以遍历顺序同插入排序,时间复杂度O(n2) class Solution { pu…

Given a string s, find the longest palindromic substring in s. You may assume that the maximum length of s is 1000. Example: Input: "babad" Output: "bab" Note: "aba" is also a valid answer. Example: Input: "cbbd" Ou…

Given a string S, find the longest palindromic substring in S. You may assume that the maximum length of S is 1000, and there exists one unique longest palindromic substring. 思路:如果不用动态规划,在两个for循环的情况下,还得依次比较i,j间的每个字符,O(n3).使用动态规划,O(n2) char* longestPa…

Given a string S, find the longest palindromic substring in S. You may assume that the maximum length of S is 1000, and there exists one unique longest palindromic substring. 这道题让我们求最长回文子串,首先说下什么是回文串,就是正读反读都一样的字符串,比如 "bob", "level", &q…

Given a string S, find the longest palindromic substring in S. You may assume that the maximum length of S is 1000, and there exists one unique longest palindromic substring. 题目的意思:输入一个字符串S,找出其最长回文串 用动态规划求解 决策变量:dp[i][j]记录从s[i]到s[j]组成的子串是否为回文. dp[i+1…

Given a , and there exists one unique longest palindromic substring. https://leetcode.com/problems/longest-palindromic-substring/ 求最大回文的长度,其实这道题比上一道有意思. 方法1 循环查询 (该方案为O(N*N*N)) public static boolean isPalindrome(String s, int start, int end) { if (((…

Given a string S, find the longest palindromic substring in S. You may assume that the maximum length of Sis 1000, and there exists one unique longest palindromic substring. == class Solution { public: string longestPalindrome(string s) { /**bool dp[…