Longest Palindromic Substring - 一题多解

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Longest Palindromic Substring - 一题多解

题意是寻找一个字符串的最大回文字串,最简单的是n3方的算法,由于字符串最大长度为1000,所以这个方法很危险而且不科学. 紧接着想到的是一个n方的算法:回文子串是从中间向两边产生的,那么对于每个字符考察这个字符往外的所有可能不就可以找到以这个字符为中心的最长回文子串了吗?当然要考虑偶数的情况,即那个中心不在字母上,而这种情况的字符又一定满足中心的两个字符是相等的.所以在第一遍寻找奇数最长回文字符的时候记录下有可能形成偶数回文字符的数组下标,统一规定为两个字母下标较小的那个. 按照以上想法编码AC…

LeetCode第[5]题(Java)：Longest Palindromic Substring 标签：String、动态规划

题目中文:求最长回文子串题目难度:Medium 题目内容: Given a string s, find the longest palindromic substring in s. You may assume that the maximum length of s is 1000. 翻译: 给定一个字符串s,找出s中最长的回文子串.你可以假设s的最大长度是1000. 什么叫回文子串? 就是字符串中,满足能正读反读都一样的子串,就是回文子串.如下所示 Input: "babad"…

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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. 答案一道动规题目,每次用当前位置columnIndex跟前面位置rowIndex去比较时,如果相同的话,就去看columnIndex- 1和r…

leetcode 第五题 Longest Palindromic Substring （java）

Longest Palindromic Substring 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. time=255ms accepted 暴力遍历 public String lon…

leetcode第五题--Longest Palindromic Substring

Problem: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. 找出最大的回文子串,回文就是指字符串倒过来和之前的一样.例如aba 倒过来还是aba.abac中的最大回文子串就是aba. 我一开始…

【LeetCode每天一题】Longest Palindromic Substring(最长回文字串)

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:…

【LeetCode算法题库】Day2：Median of Two Sorted Arrays & Longest Palindromic Substring & ZigZag Conversion

[Q4] There are two sorted arrays nums1 and nums2 of size m and n respectively. Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)). You may assume nums1 and nums2 cannot be both empty. Example 1: nums1 = […