LeetCode 260

Given an array of Numbers nums, in which exactly two elements appear only once and  all 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].

Algorithm should run in linear runtime complexity and using constant space complexity. 

This is also a bit manipulation problem. 

// The first step is to get ride of the elements which appear twice. As we know in LeetCode 136, A ^ A = 0;
vector<int> singleNumber(vector<int>& nums) {
    // we first need to check the input size.
    if(nums.size() < 2) return {};
    int result = 0;
    for(auto it : nums) {
        result ^= it;
    }
    // currently, the result is num1 ^ num2, in order to split them out, we need to find the last different bit which is '1'
    // this equation is very important, just remember it!
    int diff = result & ~(result -1);
    int tmp = 0;
    for(auto it : nums) {
         if(diff & it) {
             tmp ^= it;
        }
    }
    // A ^ B ^ A = B, thus, the return value would be
    return {tmp, result ^ tmp};
}