Python解法:680. Valid Palindrome II

class Solution(object):
    def validPalindrome(self, s):
        l = len(s)
        if l == 2:
            return True
        i = 0
        j = l - 1
        count = 0
        while(i<j):
            if s[i] != s[j]:
                i += 1
                count += 1
            else:
                j -= 1
                i += 1
        if count < 2:
            return True
        count = 0
        i = 0
        j = l - 1
        while(i<j):
            if s[i] != s[j]:
                j -= 1
                count += 1
            else:
                j -= 1
                i += 1
        if count < 2:
            return True
        return False

思路

指针i，j从字符串头尾开始比较，只要有>1个值不一样，就没有办法回文。