In computer science, the longest palindromic substring or longest symmetric factor problem is the problem of finding a maximum-length contiguous substring of a given string that is also a palindrome. For example, the longest palindromic substring of "bananas" is "anana". The longest palindromic sub...
Manacher算法代码如下: 1publicString longestPalindromeOne(String s) {23char[] cs =expandString(s);4intlen =cs.length;5int[] p =newint[len];6intid = 0;7intmax = 0;8for(inti = 1; i <len-1 ; i++){9if(i < p[id]+id){10p[i]=Math.min(p[id]+id-i, p[2*id-i]);11}...
public class Solution { /** * @param s: input string * @return: the longest palindromic substring */ public String longestPalindrome(String s) { // write your code here char[] S = s.toCharArray(); StringBuilder res = new StringBuilder("$#"); for (int i=0;i<S.length;i++){ res...
Given a string s, find the longest palindromic substring in s. You may assume that the maximum length of s is 1000. 难度:Medium 2 题目样例 Input: "babad" Output: "bab" Note: "aba" is also a valid answer. 3 题目分析 给出一个字符串s,要求求出这个字符串中的最长的回文子串。 回文串...
Longest Palindromic Substring 描述 Given a string S, find the longest palindromic substring in S. You may assume that the maximum length of …
Longest Palindromic Substring 最长的回文字串 Longest Palindromic Substring Given a string S, find the longest palindromic substring in S. You may assume that the maximum length of S class Solution { public: string longestPalindrome(string s) {...
Given a string s, find the longest palindromic(回文) substring(子串) 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" ...
1 题目叙述Given a stringS, find the longest palindromic substring inS. You may assume that the maximum length ofSis 1000, and there exists one unique longest palindromic substring.2 本体可以使用动态规划去解答，但是我用了之后没能AC，也可以使用从中间向两边延伸去查找最长回文子串。先提供第二种方式的...
5. Longest Palindromic Substring Given a strings, find the longest palindromic substring ins. You may assume that the maximum length ofsis 1000. 难度:medium sol 2 Brute Force 挨个检查所有子串,子串O(n^2)*检测所需时间O(n)=O(n^3).
String algorithmLongest common palindromic subsequenceRecently, Chowdhury et al. [5] proposed the longest common palindromic subsequence problem . It is a variant of the well-known LCS problem, which refers to finding a palindromic LCS between two strings T 1 T 1 mathContainer Loading Mathjax and...