leetcode-409.最长回文子串

题目来自leetcode
https://leetcode-cn.com/problems/longest-palindrome/

解决思路：
遍历字符串，统计每个字符出现的次数。
统计出现次数为奇数(计为odd)的字符个数
若odd=0，则最长回文子串=字符串长度
若odd>0,则最长回文子串 = 字符串长度+1-odd

class Solution:
    def longestPalindrome(self, s: str) -> int:
        l = len(s)
        d = {}
        for i in range(l):
            temp = s[i]
            if temp in d:
                d[temp] += 1
            else:
                d[temp] = 1
        
        odd = 0
        for value in d.values():
            if value % 2 == 1:
                odd += 1
        
        if odd>0:
            result = l+1-odd
        else:
            result = l
        return result