[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....
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...
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] != ...
For the purpose of this problem, we define empty string as valid palindrome. Solution 把不是数字和字母的字符先去掉,然后统计大小写,最后判断是否为回文。 Code classSolution{public:boolisPalindrome(strings){if(s.empty())returntrue;stringstr;for(charc : s) {if(isDigitAlpha(c)) {if(c >='a'...
publicclassSolution { publicbooleanisPalindrome(String s) { StringBuilder str =newStringBuilder(); for(inti=0; i<s.length(); i++) { if(s.charAt(i)>='a'&&s.charAt(i)<='z'|| s.charAt(i)>='A'&&s.charAt(i)<='Z' || s.charAt(i)>='0'&&s.charAt(i)<='9') { ...
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*...
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 + ...
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'....
For the purpose ofthisproblem, we define empty string as valid palindrome. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 第二次做法:用API 1publicclassSolution {2publicbooleanisPalindrome(String s) {3if(s ==null)returnfalse;4if(s.length() == 0)returntrue;5intl=0, r=s.length()-1;6whi...
classSolution{public:boolisPalindrome(string s){intl =0, r = s.size() -1;while(l <= r){while(!isalnum(s[l]) && l < r) l++;while(!isalnum(s[r]) && l < r) r--;if(toupper(s[l]) !=toupper(s[r]))returnfalse;