*LeetCode--Valid Palindrome II Valid Palindrome II Given a non-empty strings, you may delete at most one character. Judge whether you can make it a palindrome. Example 1: Input: "aba" Output: True Example 2: Input: "abca" Output: True Explanation: You could delete the character 'c'....
3.1 Java实现 publicclassSolution{intpos=0;publicbooleanvalidPalindrome(String s){booleanvalid=isPalindrome(s);if(!valid) {intleft=pos;intright=s.length() - pos;booleanret1=isPalindrome(s.substring(left +1, right));booleanret2=isPalindrome(s.substring(left, right -1)); valid = ret1 || ...
680. Valid Palindrome II Given a strings, returntrueif thescan be palindrome after deletingat most onecharacter from it. Example 1: Input:s = "aba"Output:true Example 2: Input:s = "abca"Output:trueExplanation:You could delete the character 'c'. Example 3: Input:s = "abc"Output:false...
func validPalindrome(s string) bool { // 定义左指针 l ,初始化为 0 l := 0 // 定义右指针 r ,初始化为 s.length - 1 r := len(s) - 1 // 当还有字符需要比较时,继续处理 for l < r { // 如果 s[l] 和 s[r] 不相等,则需要删除字符 if s[l] != s[r] { // 如果删除 左指...
LeetCode-Valid Palindrome II Description: Given a non-empty string s, you may delete at most one character. Judge whether you can make it a palindrome. Example 1: AI检测代码解析 Input: "aba" Output: True 1. 2. Example 2: AI检测代码解析...
Valid Palindrome·有效回文 秦她的菜 吉利 程序员 题目描述 英文版描述 A phrase is a palindrome if, after converting all uppercase letters into lowercase letters and removing all non-alphanumeric characters, it reads the same forward and backward. Alphanumeric characters include letters and ...
[Leetcode][python]Valid Palindrome/验证回文串 题目大意 判断一个字符串是否是回文字符串,只考虑字母和数字,并且忽略大小写。 注意点: 空字符串在这里也定义为回文串 解题思路 去掉除了数字和字母之外的字符isalnum() 都改为小写 将数组(字符串)反过来,判断是否相等...
0680 Valid Palindrome II LeetCode 力扣 Python CSDN Easy 双指针 0695 Max Area of Island LeetCode 力扣 Python CSDN Medium DFS 0700 Search in a Binary Search Tree LeetCode 力扣 Python CSDN Easy 二叉树 0701 Insert into a Binary Search Tree LeetCode 力扣 Python CSDN Medium 二叉树 0714 Best Time...
class Solution(object): def validPalindrome(self, s): def check(left, right): # 判断是否是回文数 while left < right: if s[left] != s[right]: return False left += 1 right -= 1 return True # 定义指针,一个指向开始,一个指向末尾 left, right = 0, len(s) - 1 while left < righ...
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