Leetcode: 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 t Wice. Find the two elements this appear only once.

For example:

Given nums = [1, 2, 1, 3, 2, 5], return [3, 5].

Note:
The order of the ' is ' not important. The above example, [5, 3] is also correct.
Your algorithm should run in linear runtime complexity. Could you implement it using only constant space complexity?

For the given array, except for two numbers, the other numbers appear two times and find the two numbers that occur once.

Idea: In fact, as with single number, just add a trick, first to store all the difference or the results, then this number must be the occurrence of 1 of the two number of differences or results. In the binary representation of the result, a bit with a value of 1 must be a different bit of these two numbers, such as 3 (011) and 5 (101), with two digits. This means that there must be a certain one, a number at which the value is 1 and the other is 0. So we can divide the number of the original array into two groups, and use single number to find out the two numbers.

Code:

 public class Solution {public int[] Singlenumber (int[] nums) {if (Nums = null | | nums.length = 0) {
        return null;

        int diff = 0;
        for (int num:nums) {diff ^= num;

        Diff &=-diff;

        Int[] result = new INT[2];
            for (int num:nums) {if ((num & diff) = = 0) {result[0] ^= num;
            }else{result[1] ^= num;
    } return result; }
}