266. Palindrome Permutation - Easy

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

扫一遍s，统计频率，奇数频率最多只能出现一次。用cnt表示结果，cnt += map[i]的频率%2，如果cnt > 1，则不可能是回文

time: O(n), space: O(n)

class Solution {
    public boolean canPermutePalindrome(String s) {
        HashMap<Character, Integer> map = new HashMap<>();
        for(int i = 0; i < s.length(); i++) {
            map.put(s.charAt(i), map.getOrDefault(s.charAt(i), 0) + 1);
        }

        int cnt = 0;
        for(Map.Entry<Character, Integer> entry : map.entrySet()) {
            cnt += entry.getValue() % 2;
        }
        return cnt <= 1;
    }
}

原文地址：https://www.cnblogs.com/fatttcat/p/10179670.html