Palindromic Substring Time Limit: 10000ms Memory Limit: 65536KB This problem will be judged on HDU. Original ID: 442664-bit integer IO format: %I64d      Java class name: Main In the kingdom of string, people like palindromic strings very much. They…

综合性很强的一道题目,结合manacher,后缀数组,哈希,RMQ,二分可解.基本思路是通过manacher可以找到所有可能的回文串,哈希去重,后缀数组二分找数目.最后暴力求解.需要注意kth需要为__int64. /* 4426 */ #include <iostream> #include <sstream> #include <string> #include <map> #include <queue> #include <set&…

题目描述 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. 即给定一个字符串,求它的最长回文子串的长度(或者最长回文子串). 解法一 对于一个问题,一定可以找到一个傻的可爱的暴力解法,本题的暴力解法即…

题目来自 https://leetcode.com/problems/longest-palindromic-substring/ 题目: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. 给一个字符串…

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…

问题: 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. 官方难度: Medium 翻译: 给定一个字符串S,找出最长回文子串,你可以假定字符串的最大长度为1000,并且只有一个最长回文子串. 例子:…

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 (((…

LeetCode: 5. Longest Palindromic Substring class Solution { public: //动态规划算法 string longestPalindrome(string s) { int n = s.length(); int longestBegin = 0; int maxLen = 1; bool table[1000][1000] = {false}; for (int i = 0; i < n; i++) { table[i][i] =…

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[…