[leetcode] 266. Palindrome Permutation @ python

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原题

Given a string, determine if a permutation of the string could form a palindrome.

Example 1:

Input: “code”
Output: false
Example 2:

Input: “aab”
Output: true
Example 3:

Input: “carerac”
Output: true

解法

观察字符串, 能组成回文的条件是最多有1个奇数个的字母放在回文中间, 剩下的必须是偶数个字母. 因此计算s中字母出现次数, 定义chance为1, 当字母出现奇数次时, chance减1, 当chance为负数时返回False.
Time: O(n)
Space: O(1)

代码

class Solution(object):
    def canPermutePalindrome(self, s):
        """
        :type s: str
        :rtype: bool
        """
        count = collections.Counter(s)
        chance = 1
        for char in count:
            if count[char]%2 != 0:
                chance -= 1
                if chance < 0:
                    return False
        return True