Single Number III leetcode

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].

与single number1 类似 每个重复的数都出现两次, 假设不重复的数字为a, b. 用^遍历所有数组就会得到一个a^b的结果c. 我们知道在c中, 每一位1说明在这一位上a与b是不同的. 
      3 (0011) xor 
      5 (0101)= 
     6 (0110) 
我们只需要找到这1位, 然后把数组按照这一位是1还是这一位是0分为两个数组 然后分别做^遍历 就会在每个组中找到一个独立的单身狗了.

public class Solution {
    public int[] singleNumber(int[] nums) {
        int[] res = new int[2];
        if (nums == null || nums.length == 0) {
            return res;
        }
        int tem = 0;
        for (int c : nums) {
            tem ^= c;
        }
        int pre = 0;
        while (tem!= 0) {
            pre = tem;
            tem &= tem - 1;
        }
        for (int n : nums) {
            if ((n & pre) == 0) {
                res[0] ^= n;
            } else {
                res[1] ^= n;
            }
        }
        return res;
    }
}