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:

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?

解題思路

位運算。與Single Number的區別在於首先根據所有元素的亦或結果的某一位是否為1進行分組，然後分別找出每組中的Single Number。

實現代碼

Java：

// Runtime: 2 ms
public class Solution {
    public int[] singleNumber(int[] nums) {
        int xor = 0;
        for (int num : nums) {
            xor ^= num;
        }
        int bit = xor & (~(xor - 1));

        int single[] = new int[2];
        for (int num : nums) {
            if ((num & bit) != 0) {
                single[0] ^= num;
            } else {
                single[1] ^= num;
            }
        }

        return single;
    }
}