*Single Number

Given an array of numbers nums, in which exactly two elements appear only once and all the other elements appear exactly twice. Find the two elements that appear only once.

For example:

Given nums = [1, 2, 1, 3, 2, 5], return [3, 5].

Note:

  1. The order of the result is not important. So in the above example, [5, 3] is also correct.
  2. Your algorithm should run in linear runtime complexity. Could you implement it using only constant space complexity?

 

Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

 

 1 /**
 2  * 本代码由九章算法编辑提供。没有版权欢迎转发。
 3  * - 九章算法致力于帮助更多中国人找到好的工作,教师团队均来自硅谷和国内的一线大公司在职工程师。
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 5  * - 更多详情请见官方网站:http://www.jiuzhang.com/
 6  */
 7 
 8 public class Solution {
 9     /**
10      * @param A : An integer array
11      * @return : Two integers
12      */
13     public List<Integer> singleNumberIII(int[] A) {
14         int xor = 0;
15         for (int i = 0; i < A.length; i++) {
16             xor ^= A[i];
17         }
18         
19         int lastBit = xor - (xor & (xor - 1));
20         int group0 = 0, group1 = 0;
21         for (int i = 0; i < A.length; i++) {
22             if ((lastBit & A[i]) == 0) {
23                 group0 ^= A[i];
24             } else {
25                 group1 ^= A[i];
26             }
27         }
28         
29         ArrayList<Integer> result = new ArrayList<Integer>();
30         result.add(group0);
31         result.add(group1);
32         return result;
33     }
34 }

 

posted @ 2015-10-08 04:30  Hygeia  阅读(138)  评论(0编辑  收藏  举报