260. Single Number III

Description

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?

Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

Solution

Two-pass XOR, time O(n), space O(1)

先取XOR，然后找到XOR中二进制为1的一位，然后根据它将nums分成两个group，然后分别取XOR即可。

class Solution {
    public int[] singleNumber(int[] nums) {
        int xor = 0;
        for (int n : nums) {
            xor ^= n;
        }
        
        // find one different digit
        int mask = 1;
        for (int i = 0; i < 32 && (xor & mask) == 0; ++i) {
            mask <<= 1;
        }
        
        int[] res = new int[2];
        // divide nums into two groups base on the different digit
        for (int n : nums) {
            if ((n & mask) == 0) {
                res[0] ^= n;
            } else {
                res[1] ^= n;
            }
        }
        
        return res;
    }
}