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.

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?

Solution: runtime complexity is O(n), constant space.
"we notice that A and B must be different at some bit at position t in their binary representations. So if we divide the set of numbers into 2 set, one is the set of all the numbers that have the same bit at position t as A, the other is the set of all numbers that have the same bit at position t as B. These 2 sub-sets have a special characteristic: all numbers appear 2 times, except 1. This bring us to the Single Number problem."
Reference: http://traceformula.blogspot.com/2015/09/single-number-iii-leetcode.html

public class Solution {
    public int[] singleNumber(int[] nums) {
        int A = 0;
        int B = 0;
        int AXORB = 0;
        
        for (int n : nums) {
            AXORB ^= n;
        }
        
        int lastBit = (AXORB & (AXORB - 1)) ^ AXORB;  // find the last bit that A differs from B
        
        for (int n : nums) {  
            // based on the last bit, group the items into groupA (include A) and groupB
            if ((lastBit & n) == 0) {
                A ^= n;
            } else {
                B ^= n;
            }
        }
        
        return new int[]{A, B};
    }
}