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; } }
266. Palindrome Permutation - Easy
原文：https://www.cnblogs.com/fatttcat/p/10179670.html