Single Number III

class Solution {
public:
    /**
     * @param A : An integer array
     * @return : Two integers
     */
    vector<int> singleNumberIII(vector<int> &A) {
        // write your code here
        int a_xor_b = 0;
        int a = 0;
        int b = 0;
        
        //Get xor result of the two single numbers. 
        //The bit which is 1 indicates at that bit A and B are different
        //The bit which is 0 indicates at that bit A and B are the same
        for (int &i : A){
            a_xor_b ^= i;
        }
        
        
        /***
         *  a_xor_b - 1: From the rightest(last) bit
         *               1) if the bit is 1, set that bit to 0, and keep the rest unchange and exit. 
         *               2) if the bit is 0, set that bit to 1, and check previous bit until 1)
         * ~(a_xor_b -1): The changed bit will be changed back, unchanged bit will be changed to opposite. 
         * a_xor_b & ~(a_xor_b -1): 1) unchanged bit will all be set to 0 (now they are different)
         *                          2) changed bit will be kept as original. 
         * The result of this opperation will provide us a number, 
         * which from the right(last), the first 1 will be kept and rest of bits will be set to 0,
         * comparing with a_xor_b;
         * eg. a_xor_b = 3(0011), (a_xor_b-1) = 2(0010), ~(a_xor_b-1) = (1101), a_xor_b & ~(a_xor_b -1) = (0001)
         *     a_xor_b = 5(0101), (a_xor_b-1) = 4(0100), ~(a_xor_b-1) = (1011), a_xor_b & ~(a_xor_b -1) = (0001)
         *     a_xor_b = 4(0100), (a_xor_b-1) = 3(0011), ~(a_xor_b-1) = (1100), a_xor_b & ~(a_xor_b -1) = (0100)
         *     a_xor_b = 6(0110), (a_xor_b-1) = 5(0101), ~(a_xor_b-1) = (1010), a_xor_b & ~(a_xor_b -1) = (0010)
         ***/
        int first_different_bit_from_right = a_xor_b & ~(a_xor_b - 1);
        
        /***
         * Second way:
         * // Get the last bit where 1 occurs. 
         * int first_different_bit_from_right = x_xor_y & -x_xor_y;
         * -x = ~x+1. Similar reason. 
         ***/
        
        
        //Divided the whole array to two groups,
        //The first group contains the numver which at particular bit, the bit value is 1;
        //The particular bit is the bit which is the rightest bit that a and b are different.
        //The second group contains number which at the particular bit, the bit value is 0;
        //Since at that bit, a and b are different,a and b must assign to different group.
        //The rest member of each group is even(same number will be sign to same group). 
        //If we XOR each group, the result will be the leftover one single number in each group
        
        for(int &i : A){
            if (i & first_different_bit_from_right ){
                a ^= i;
            } else {
                b ^= i;
            }
        }
        
        return {a, b};
        
        
    }
};