Single Number III(LintCode)

Single Number III

Given 2*n + 2 numbers, every numbers occurs twice except two, find them.

Example

Given [1,2,2,3,4,4,5,3] return 1 and 5

Challenge

O(n) time, O(1) extra space.

 

 

可耻的百度了。。点这里

还是好理解的。当然不一定要用最低位的1,任意找一个1即可。不过有lowbit = xor & -xor 这个公式当然是这个比较快。。

 1 public class Solution {
 2     /**
 3      * @param A : An integer array
 4      * @return : Two integers
 5      */
 6     public List<Integer> singleNumberIII(int[] A) {
 7         List<Integer> list = new ArrayList<Integer>();
 8         
 9         int x = 0;
10         for(int a : A) {
11             x = x ^ a;
12         }
13         int lowbit = x & (-x);
14         int a = 0;
15         int b = 0;
16         for(int y : A) {
17             if((y & lowbit) == 0) {
18                 a = a ^ y;
19             }else {
20                 b = b ^ y;
21             }
22         }
23         
24         
25         list.add(a);
26         list.add(b);
27         return list;
28     }
29 }
View Code

 

posted @ 2015-12-08 23:56 -.-| 阅读(...) 评论(...) 编辑 收藏