Given a list of unique words. Find all pairs of distinct indices (i, j) in the given list, so that the concatenation of the two words, i.e. words[i] + words[j] is a palindrome. Example 1:Given words = ["bat", "tab", "cat"]Ret…

Given a string, determine if a permutation of the string could form a palindrome. For example,"code" -> False, "aab" -> True, "carerac" -> True. Hint: Consider the palindromes of odd vs even length. What difference d…

给定一组独特的单词, 找出在给定列表中不同 的索引对(i, j),使得关联的两个单词,例如:words[i] + words[j]形成回文.示例 1:给定 words = ["bat", "tab", "cat"]返回 [[0, 1], [1, 0]]回文是 ["battab", "tabbat"]示例 2:给定 words = ["abcd", "dcba", &q…

Given a string S, you are allowed to convert it to a palindrome by adding characters in front of it. Find and return the shortest palindrome you can find by performing this transformation. For example: Given "aacecaaa", return "aaacecaaa&qu…

Given a string s, you are allowed to convert it to a palindrome by adding characters in front of it. Find and return the shortest palindrome you can find by performing this transformation. Example 1: Input: "aacecaaa" Output: "aaacecaaa&quo…

Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases. For example, "A man, a plan, a canal: Panama" is a palindrome. "race a car" is not a palindrome. Note: Have you consider that…

class Solution { public: bool isPalindrome(string s) { if(s=="") return true; ) return true; //单个字符,对称 char *p,*q; p=&s[]; //p指向开头 q=&s[s.length()-]; //q指向末尾 while(p!=q){ //测试字符串里是否有字母或数字,若没有,则立刻返回 '&&*p<'A') || (*p>'Z'&…

LeetCode:验证回文串[125] 题目描述 给定一个字符串,验证它是否是回文串,只考虑字母和数字字符,可以忽略字母的大小写. 说明:本题中,我们将空字符串定义为有效的回文串. 示例 1: 输入: "A man, a plan, a canal: Panama" 输出: true 示例 2: 输入: "race a car" 输出: false 题目分析 很明显这是双指针问题,回文判断的方式有很多种,最简单的是将原字符串逆序后判断两者是否相同.但是显然这里有干扰…

目录 # 前端与算法 leetcode 125. 验证回文串 题目描述 概要 提示 解析 解法一:api侠 解法二:双指针 算法 传入测试用例的运行结果 执行结果 GitHub仓库 查看更多 # 前端与算法 leetcode 125. 验证回文串 题目描述 给定一个字符串,验证它是否是回文串,只考虑字母和数字字符,可以忽略字母的大小写. 说明:本题中,我们将空字符串定义为有效的回文串. 示例 1: 输入: "A man, a plan, a canal: Panama" 输出: tru…

CF932G Palindrome Partition(回文自动机) Luogu 题解时间 首先将字符串 $ s[1...n] $ 变成 $ s[1]s[n]s[2]s[n-1]... $ 就变成了求将字符串全部划分为偶回文串的方案数. 建回文树大力跳$ fail $ 直接 $ dp $ 的复杂度是十分优秀的 $ O ( n ^ {2} ) $. 优化不容易想到. 考虑字符串上第 $ j $ 位为结尾的所有回文子串,毫无疑问它们在树上是一条链. 但它有个更重要的性质. 其中所有长度 $ > j…