[NeetCode/Python] Is Palindrome

문제: https://neetcode.io/problems/is-palindrome

NeetCode

 

[내 풀이]

class Solution:
    def isPalindrome(self, s: str) -> bool:
        s = ''.join(x for x in s if x.isalnum()).lower()

        for i in range(len(s)//2):
            l = 0 + i
            r = len(s) - 1 - i

            if s[l] == s[r]:
                pass
            else:
                return False

        return True

 

python은 코테 언어다....

 

[솔루션 풀이]

class Solution:
    def isPalindrome(self, s: str) -> bool:
        l, r = 0, len(s) - 1

        while l < r:
            while l < r and not self.alphaNum(s[l]):
                l += 1
            while r > l and not self.alphaNum(s[r]):
                r -= 1
            if s[l].lower() != s[r].lower():
                return False
            l, r = l + 1, r - 1
        return True
    
    def alphaNum(self, c):
        return (ord('A') <= ord(c) <= ord('Z') or 
                ord('a') <= ord(c) <= ord('z') or 
                ord('0') <= ord(c) <= ord('9'))

 

시간 복잡도

n은 문자열의 길이

O(n/2)