LeetCode 솔루션

[Easy - wk6 - Q1] 125. Valid Palindrome

Chloe 쪽지보내기 메일보내기 자기소개 아이디로 검색 전체게시물 작성
작성일 2022.06.07 01:05

6,356 조회
3 댓글
0 추천
0 비추천
목록 게시글 쓰기

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 numbers.

Given a string s, return true if it is a palindrome, or false otherwise.

Example 1:

Input: s = "A man, a plan, a canal: Panama"
Output: true
Explanation: "amanaplanacanalpanama" is a palindrome.

Example 2:

Input: s = "race a car"
Output: false
Explanation: "raceacar" is not a palindrome.

Example 3:

Input: s = " "
Output: true
Explanation: s is an empty string "" after removing non-alphanumeric characters.
Since an empty string reads the same forward and backward, it is a palindrome.

Constraints:

1 <= s.length <= 2 * 105
s consists only of printable ASCII characters.

#leetcode, #easy, #문제풀이, #facebook, #amazon, #microsoft, #apple, #bloomberg, #linkedin, #wayfair, #spotify, #adobe, #servicenow, #jpmorgan, #american express, #two pointers, #string

익명
작성일 2022.06.08 20:41

class Solution:
def isPalindrome(self, s: str) -> bool:

s2 = [x.lower() for x in s if x.isalpha() or x.isnumeric()]
left = 0
right = len(s2) - 1

while left < right:
if s2[left] != s2[right]:
return False
left += 1
right -= 1

return True

익명
작성일 2022.06.10 15:51

class Solution {
    
    public boolean isPalindrome(String s) {
        // two pointer
        for(int i=0, j=s.length()-1; i<j; i++, j--){
            while(i<j && !Character.isLetterOrDigit(s.charAt(i))){
                i++;
            }
            while(i<j && !Character.isLetterOrDigit(s.charAt(j))){
                j--;
            }
            
            if(Character.toLowerCase(s.charAt(i)) != Character.toLowerCase(s.charAt(j))){
                return false;
            }
        }
        
        return true;
    }
    
    
}

익명
작성일 2022.06.10 15:58

Runtime: 93 ms, faster than 19.64% of Python3 online submissions for Valid Palindrome.
Memory Usage: 14.7 MB, less than 43.06% of Python3 online submissions for Valid Palindrome.

class Solution:
    def isPalindrome(self, s: str) -> bool:
        new_s = ""
        for char in s:
            if ord('a') <= ord(char.lower()) <=ord('z'): new_s += char.lower()
            elif char.isdigit(): new_s += char
        return new_s[::-1] == new_s