Verification palindrome string II of leetcode simple question

subject

Given a non empty string s, Delete at most one character . Determine whether it can be a palindrome string .

Example 1:

Input : s = “aba”
Output : true
Example 2:

Input : s = “abca”
Output : true
explain : You can delete c character .
Example 3:

Input : s = “abc”
Output : false

Tips :

1 <= s.length <= 10^5
s It's made up of lowercase letters

source ： Power button （LeetCode）

Their thinking

   Whether it is palindrome is an old question , The common method can be traversal or stack , Here we use traversal . For the case that the string cannot be palindrome, we only need to check two cases , That is, the first half and the second half .

class Solution:
    def validPalindrome(self, s: str) -> bool:
        def Palindrome(x):
            for i in range(len(x)//2):
                if x[i]!=x[-i-1]:
                    return i
            return None
        temp=Palindrome(s)
        if temp==None:
            return True
        else:
            s1=s[0:temp]+s[temp+1:]
            s2=s[0:len(s)-1-temp]+s[len(s)-temp:]
            return False if Palindrome(s1)!=None and Palindrome(s2)!=None else True