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.

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

The order of the result is not important. So in 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?

• 考虑使用位运算
  -那数组分为两组，每个组只有一个出现一次的数字，单独进行异或处理，考虑用全部异或的方式

• 假设得到异或的结果是A，由于两个数字不同。A肯定有一个位是1，找到那个位，然后用来把这个数组分为两组
• 其中一个组的异或结果是B，输出【B，A ^ B】

public class Solution {
    public int[] singleNumber(int[] nums) {
        int xOne = 0;
        for (int x : nums) {
            xOne ^= x;
        }

        // 获取第一个位1的索引
        int k = 0;
        for (k = 0; k < Integer.SIZE; ++ k) {
            if (((xOne >>> k) & 1) == 1) {
                break;
            }
        }

        int xTwo = 0;
        for (int x : nums) {
            if (((x >>> k) & 1) == 1) {
                xTwo ^= x;
            }
        }
        return new int[] {xTwo, xOne ^ xTwo};
    }
}