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…

LeetCode: Palindrome Partition Given a string s, partition s such that every substring of the partition is a palindrome. Return all possible palindrome partitioning of s. For example, given s = "aab", Return [ ["aa","b"], [&q…

Given a string s, partition s such that every substring of the partition is a palindrome. Return the minimum cuts needed for a palindrome partitioning of s. For example, given s ="aab",Return1since the palindrome partitioning["aa",&quo…

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…

LeetCode:Palindrome Partitioning 题目如下:(把一个字符串划分成几个回文子串,枚举所有可能的划分) Given a string s, partition s such that every substring of the partition is a palindrome. Return all possible palindrome partitioning of s. For example, given s = "aab",Return [ […

Given a string s, partition s such that every substring of the partition is a palindrome. Return the minimum cuts needed for a palindrome partitioning of s. For example, given s = "aab",Return 1 since the palindrome partitioning ["aa",…

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…

Given a string s, partition s such that every substring of the partition is a palindrome. Return all possible palindrome partitioning of s. Example Given s = "aab", return: [ ["aa","b"], ["a","a","b&q…

LeetCode: Palindrome 回文相关题目汇总 LeetCode: Palindrome Partitioning 解题报告 LeetCode: Palindrome Partitioning II 解题报告 Leetcode:[DP]Longest Palindromic Substring 解题报告 LeetCode: Valid Palindrome 解题报告…

[1]LeetCode 136 Single Number 题意:奇数个数,其中除了一个数只出现一次外,其他数都是成对出现,比如1,2,2,3,3...,求出该单个数. 解法:容易想到异或的性质,两个相同的数异或为0,那么把这串数从头到尾异或起来,最后的数就是要求的那个数. 代码如下: class Solution { public: int singleNumber(vector<int>& nums) { ; ;i<nums.size();i++) sum ^= nums[i…