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.

Example:

Input:  [1,2,1,3,2,5]
Output: [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?

和136不同的是，這裡有2個只出現一次的數字。還是先把所有數字xor一遍，出來的結果是這兩個數的xor，即 tmp = a ^ b，現在需要分離這兩個數。可以任選其中的一個為1的位，比如tmp & -tmp，是最右的1，然後把tmp和所有數字相與(&)，通過這一步可以把答案的兩個數分到兩個組，然後再對兩個小組分別異或，就可以得到這兩個數

time: O(N), space: O(1)

class Solution {
    public int[] singleNumber(int[] nums) {
        int[] res = new int[2];
        int tmp = 0;
        for(int k : nums) {
            tmp ^= k;
        }
        
        tmp &= -tmp;
        
        for(int k : nums) {
            if((k & tmp) == 0) res[0] ^= k;
            
 
else res[1] ^= k;
        }
        return res;
    }
}

reference: http://www.cnblogs.com/grandyang/p/4741122.html