Single Number III

标签：return   http   example   iii   imp   pass   not   turn   runtime   

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 public class Solution {
 2     public int[] singleNumber(int[] nums) {
 3         // first pass to get xor result of all numbers
 4         int temp = 0;
 5         for (int num : nums)
 6             temp ^= num;
 7         
 8         // get the set bit
 9         temp &= -temp;
10         
11         int[] result = {0, 0};
12         for(int num : nums) {
13             if ( (temp & num) == 0) {
14                 result[0] ^= num;
15             } else {
16                 result[1] ^= num;
17             }
18         }
19         return result;
20     }
21 }

Single Number III

标签：return   http   example   iii   imp   pass   not   turn   runtime   

原文地址：http://www.cnblogs.com/amazingzoe/p/6416390.html