题目: https://leetcode.com/problems/palindrome-permutation/
难度 : Easy
思路:
hint 已经提示的很明显。数单字个数来处理
AC代码
class Solution(object):
def canPermutePalindrome(self, s):
"""
:type s: str
:rtype: bool
"""
lookup = {}
for char in s:
lookup[char] = lookup.get(char,0) + 1
res = 0
for char, cnt in lookup.items():
if cnt % 2 == 0 :
continue
else:
res += 1
return res < 2