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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