Day024 - 125. Valid Palindrome

125. Valid Palindrome

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.

내 풀이

import re

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

        s = re.findall("[a-zA-Z0-9]", s) # .isalnum()
        s = "".join(s)
        s = s.lower()

        if len(s) == 1: return False

        # 1. 2-pointer를 이용한 방법
        # l = 0
        # r = len(s)-1

        # while l < r:
        #     print(l, r)
        #     if s[l] == s[r]:
        #         l += 1
        #         r -= 1
        #     else:
        #         return False
        # return True


        # 2. list slicing을 이용한 방법
        l_str = s[:len(s)//2]
        r_str = s[len(s)//2:][::-1]
        
        return l_str == r_str[:len(l_str)]

# Time Complexity : \(O(N)\)