Leetcode: Palindrome Permutation

最新推荐文章于 2021-06-09 09:25:33 发布

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最新推荐文章于 2021-06-09 09:25:33 发布

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原文链接：http://www.cnblogs.com/EdwardLiu/p/5069359.html

本文介绍了一种算法，用于判断给定字符串是否可以通过重新排列形成回文串。通过统计字符出现频率的方法来实现这一目标，并给出了具体的Java实现代码。

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

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

Hint:

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

 1 public class Solution {
 2     public boolean canPermutePalindrome(String s) {
 3         if (s==null || s.length()==0) return true;
 4         int[] count = new int[256];
 5         for (int i=0; i<s.length(); i++) {
 6             char c = s.charAt(i);
 7             count[(int)(c - '\0')]++;
 8         }
 9         int countOdd = 0;
10         for (int i=0; i<256; i++) {
11             if (count[i]%2 == 1) countOdd++;
12         }
13         if (s.length()%2==0 && countOdd==0) return true;
14         if (s.length()%2==1 && countOdd==1) return true;
15         return false;
16     }
17 }