[Leetcode] 125. Valid Palindrome

題目 : 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.

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.

1.

front : 從index=0開始數的索引值

back : 字串的最尾端往前數的索引值

isalnum() : 用來判斷該字串/字元是否由數字和字母所組成

lower() : 將字串/字元轉成小寫字母

用front和back這2個索引值一一查看該字串是否回文

如果要比對的字元不是數字或字母 -> 索引值+1/-1

用lower()將字母轉成小寫，如果s[front]和s[back]不一樣，回傳 False

全部比對完回傳 True

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

        while(front<back):
            while front<back and not s[front].isalnum():
                front +=1
            
            while front<back and not s[back].isalnum():
                back -=1

            if s[front].lower() != s[back].lower():
                return False
            front += 1
            back -= 1
            
        return True