LeetCode:Valid Palindrome 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. "...(leetcode)Valid Palindrome Question Given a string, determine if it is a ...
class Solution: def validPalindrome(self, s: str) -> bool: # 定义左指针 l ,初始化为 0 l: int = 0 # 定义右指针 r ,初始化为 s.length - 1 r: int = len(s) - 1 # 当还有字符需要比较时,继续处理 while l < r: # 如果 s[l] 和 s[r] 不相等,则需要删除字符 if s[l] != ...
class Solution { public boolean isPalindrome(String s) { StringBuffer sgood = new StringBuffer(); int length = s.length(); for (int i = 0; i < length; i++) { char ch = s.charAt(i); if (Character.isLetterOrDigit(ch)) { sgood.append(Character.toLowerCase(ch)); } } String...
1publicclassSolution {2publicbooleanisPalindrome(String s) {3intlen=s.replaceAll("[^a-zA-Z0-9]","").length();4String tmp=s.replaceAll("[^a-zA-Z0-9]","").toLowerCase();5if(len==0)returntrue;6for(inti=0;i<len/2;i++){7if(tmp.charAt(i)!=tmp.charAt(len-1-i))returnfalse...
"race a car"isnota palindrome. 代码分析 public class Solution { //判断回文字符串的基本思想:从双向遍历,每一步都同时动,若有一方不符合条件,则另一方退回上位置,下个循环再同时进行 //代码优点:1、原生函数 2、for循环遍历,两个变量 public boolean isPalindrome(String s) { ...
isValid(s[right])) { right--; continue; } if (s[left] === s[right]) { left++; right--; } else { break; } } return right <= left;};C++ Code:class Solution {public: bool isPalindrome(string s) { if (s.empty()) return true; const char*...
680. Valid Palindrome II 680. 验证回文字符串 Ⅱ 给定一个非空字符串 s,最多删除一个字符。判断是否能成为回文字符串。 示例 1: 示例 2: 注意: 字符串只包含从 a-z 的小写字母。字符串的最大长度是50000。 解法一 解法二 解法一 函数isPalindrome判断字符串s是否为回文串; 主函数validPalindrome使用...
class Solution { public boolean validPalindrome(String s) { int begin = 0; int end = s.length() - 1; while (begin < s.length() && end >= 0 && s.charAt(begin) == s.charAt(end)) { begin++; end--; } return isPalindrome(s, begin, end - 1) || isPalindrome(s, begin + ...
class Solution(object): def isPalindrome(self, s): """ :type s: str :rtype: bool """ ss = [] # 这里若用string也可以,但是大数据会TLE for char in s.lower(): if char.isalnum(): ss.append(char) return ss == ss[::-1] ...
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'....