[LeetCode] Valid Palindrome, Solution 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 the string ...
1classSolution(object):2defisPalindrome(self, x):3"""4:type x: int5:rtype: bool6"""7x2 = str(x)8ifx2 == x2[::-1]:9returnTrue10else:11returnFalse 一个比较精简的代码 运行时间打败了97%的代码 但是很占内存
C++代码如下: classSolution{public:intlongestPalindrome(string s){ unordered_map<char,int> count;for(charc : s) ++count[c];intres =0;boolhasOne =false;for(autod : count) {if(d.second %2==0) res += d.second;else{ res += d.second -1; hasOne =true; } }if(hasOne) ++res;ret...
在Leetcode 336. Palindrome Pairs问题中,如何利用哈希表来优化查找过程? 文章作者:Tyan 博客:noahsnail.com | CSDN | 简书 1. Description 2. Solution 解析:Version 1简单,容易理解,但超时。Version 2是将字符串分为两部分,前半部分和后半部分,如果两部分有一部分是回文子串,则寻找另一部分的对应的回文字符串...
680. Valid Palindrome II* 680. Valid Palindrome II* https://leetcode.com/problems/valid-palindrome-ii/ 题目描述 Given a non-empty string s, you may delete at most one character. Judge whether you can make it a palindrome. Example ......
[leetcode] 1616. Split Two Strings to Make Palindrome Description You are given two strings a and b of the same length. Choose an index and split both strings at the same index, splitting a into two strings: aprefix and asuffix where a = aprefix + asuffix, and splitting b into two...
Leetcode-Easy 234. Palindrome Linked List 描述: 判断一个单链表是否左右对称 代码语言: 代码运行次数: # Definitionforsingly-linked list.#classListNode:# def__init__(self,x):# self.val=x # self.next=NoneclassSolution(object):defisPalindrome(self,head):""":type head:ListNode:rtype:bool"""...
https://leetcode.com/problems/can-make-palindrome-from-substring/ https://leetcode.com/problems/can-make-palindrome-from-substring/discuss/372044/Short-and-fast-C%2B%2B-prefix-xor-solution-beats-100 LeetCode All in One 题目讲解汇总(持续更新中...)...
本题贪心解法的思路是找到最短的相等前后缀,将字符串拆分为[前缀,中间字符串,后缀]的形式,则答案为2+中间字符串的回文段数。从而我们可以递归的求解本题,采用朴素的字符串匹配时最坏时间复杂度为O(n2)。关于算法的正确性,国际版中给出了证明。这里对Lee的证明进行翻译,并对第二种情况做出补充。情况一:短前缀...
class Solution { int seg = 0; public int longestDecomposition(String text) { find(text); //最后一段统一加1 return seg+1; } private void find(String text) { //如果最后刚好为空,全部匹配上,先减1,因为后面会统一加1 if("".equals(text)) { seg--; return; } int l = 0; int r = ...