Grind75 - 125. Valid Palindrome 파이썬

125. Valid Palindrome

Easy

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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.

 

나의 코드

class Solution:
    def isPalindrome(self, s: str) -> bool:
        word = ""
        for i in s:
            if i.isalpha() or i.isdigit():
                word += i
        word = word.lower()
        if len(word) == 0: return True
        if len(word) % 2 != 0:
            A = word[:len(word) // 2]
            B = word[(len(word)//2)+1:][::-1]
            if A == B: return True
            else: return False
        if len(word) % 2 == 0:
            A = word[:len(word) // 2]
            B = word[(len(word)//2):][::-1]
            if A == B: return True
            else: return False