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算法笔记: leetcode#266 palindrome permutation

admin 11月 12, 2020 0

问题描述

解法

分析

Python 实现

class :
    def canPermutePalindrome(self, s):
        """
        :type s: str
        :rtype: bool
        """
        left = set()
        for c in s:
            if c in left:
                left.remove(c)
            else:
                left.add(c)
        return len(left) < 2

Java 实现

class  {
    public boolean canPermutePalindrome(String s) {
        Set<Character> left = new HashSet<>();
        
        for(int i = 0; i < s.length(); i++){
            if(left.contains(s.charAt(i))){
                left.remove(s.charAt(i));
            }else {
                left.add(s.charAt(i));
            }
        }
        
        return left.size() < 2;
    }
}

时间复杂度

O(n).

空间复杂度

O(1).

链接

266. Palindrome Permutation
(English version) Algorithm Notes: Leetcode#266 Palindrome Permutation