260. Single Number III

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，这里有2个只出现一次的数字。还是先把所有数字xor一遍，出来的结果是这两个数的xor，即 tmp = a ^ b，现在需要分离这两个数。可以任选其中的一个为1的位，比如tmp & -tmp，是最右的1，然后把tmp和所有数字相与(&)，通过这一步可以把答案的两个数分到两个组，然后再对两个小组分别异或，就可以得到这两个数

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