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
题目链接 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. 求字符串的最长回文子串 算法1:暴力解法,枚举所有子串,对每个子串判断是否为回文,复杂度为O(n^3) 算法2:删除暴力解法中有很多重复的判
1297. Palindrome Time Limit: 1.0 secondMemory Limit: 16 MB The “U.S. Robots” HQ has just received a rather alarming anonymous letter. It states that the agent from the competing «Robots Unlimited» has infiltrated into “U.S. Robotics”. «U.S. Robots» s
原文链接:英文版链接 首先,我们将字符串S中插入符号“#”转化成另一个字符串T. 比如:S = "abaaba",T = “#a#b#a#a#b#a#”. 为了找到最长回文字串,我们需要围绕Ti进行扩展,Ti-d...Ti+d是一个回文,很明显d是围绕Ti形成的回文的长度. 将每个回文的长度用数组P存起来,这样,P[i]就代表围绕Ti的回文长度,最长回文字串将会是P中的最大元素. 用上面的例子,我们得到的P的结果是(从左至右): T = # a # b # a # a # b # a
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. 翻译: 找出字符串s中最长的回文子串,字符串s的最长是1000,假设存在唯一的最长回文子串 法一:直接暴力破解 O(N3)的时间复杂度,运行超