Palindrome Permutation 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 le…

Given a string s, return all the palindromic permutations (without duplicates) of it. Return an empty list if no palindromic permutation could be form. For example: Given s = "aabb", return ["abba", "baab"]. Given s = "a…

原题链接在这里:https://leetcode.com/problems/palindrome-permutation-ii/ 题目: Given a string s, return all the palindromic permutations (without duplicates) of it. Return an empty list if no palindromic permutation could be form. For example: Given s = "aabb&…

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…

原题链接在这里:https://leetcode.com/problems/palindrome-permutation/ 题目: Given a string, determine if a permutation of the string could form a palindrome. For example,"code" -> False, "aab" -> True, "carerac" -> True. 题解: 看…

Palindrome Permutation I 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…

题目一, 题目二 思路 1. 第一遍做时就参考别人的, 现在又忘记了 做的时候使用的是二维动态规划, 超时加超内存 2. 只当 string 左部分是回文的时候才有可能减少 cut 3. 一维动规. 令 cuts[i] 表示string[i, string.size()] 所需的切割数, 那么 状态转移方程为 cuts[i] = min(cuts[j]+1) j > i && string[i, j] is palindrome 时间复杂度上仍是 o(n*n), 但更新 cuts 的…

Given a string s, return all the palindromic permutations (without duplicates) of it. Return an empty list if no palindromic permutation could be form. For example: Given s = "aabb", return ["abba", "baab"]. Given s = "a…

266.Palindrome Permutation https://www.cnblogs.com/grandyang/p/5223238.html 判断一个字符串的全排列能否形成一个回文串. 能组成回文串,在字符串长度为偶数的情况下,每个字符必须成对出现:奇数的情况下允许一个字符单独出现,其他字符都必须成对出现.用一个set对相同字符进行加减即可. class Solution { public: /** * @param s: the given string * @return: if…

Problem: Given a string s, return all the palindromic permutations (without duplicates) of it. Return an empty list if no palindromic permutation could be form. For example: Given s = "aabb", return ["abba", "baab"]. Given s…