266. Palindrome Permutation

题目：

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

For example,
"code" -> False, "aab" -> True, "carerac" -> True.

Hint:

    1. Consider the palindromes of odd vs even length. What difference do you notice?
    2. Count the frequency of each character.
    3. If each character occurs even number of times, then it must be a palindrome. How about character which occurs odd number of times?

链接： http://leetcode.com/problems/palindrome-permutation/

3/5/2017

要学会几种遍历hashmap的方法。

 1 public class Solution {
 2     public boolean canPermutePalindrome(String s) {
 3         HashMap<Character, Integer> h = new HashMap<Character, Integer>();
 4         
 5         for(int i = 0; i < s.length(); i++) {
 6             if (h.containsKey(s.charAt(i))) h.put(s.charAt(i), h.get(s.charAt(i)) + 1);
 7             else h.put(s.charAt(i), 1);
 8         }
 9         int oddCount = 0;
10         boolean isEvenLength = (s.length() % 2 == 0);
11         for (HashMap.Entry<Character, Integer> entry: h.entrySet()) {
12             if (entry.getValue() % 2 != 0) {
13                 if (isEvenLength) return false;
14                 else oddCount++;
15             }
16         }
17         return oddCount <= 1;
18     }
19 }

 

posted @ 2017-03-06 06:14  panini  阅读(130)  评论(0)  编辑  收藏  举报