Leetcode Single Number III problem solution

def singleNumber(nums):
    a = set(nums)
    b = a.copy()
    for n in nums:
        if n in a:
            a.remove(n)
            continue
        if n in b: b.remove(n)
    return list(b)

class Solution {
    public int[] singleNumber(int[] nums) {
        int a = 0;
        int b = 0;
        int xor = 0;
        
        //get all xor
        for(int n : nums) xor = xor ^ n;
        
        //discover any set bit on the result, I'm using the first one
        int firstSetBit = 0;
        while(firstSetBit < 32) if ( ((xor>>firstSetBit)&1) == 1 ) break; else firstSetBit++;
        
        //put numbers with the corresponding bit set to zero into one variable
        //and if it is set to one, put it into the other variable
        for(int n : nums) {
            if ( ((n>>firstSetBit)&1) == 0 ) a ^= n;
            else b ^= n;
        }
        
        return new int[]{a,b};
    }
}

public:
    vector singleNumber(vector& nums) {
        vector op;
        long long int xorpair=0;
        for(auto &i:nums)
        {
            xorpair^=i;
        }
        long long int rsbm=xorpair&(-xorpair);
        long long int on=0;
        long long int off=0;
        for(auto &i:nums)
        {
            if(i&rsbm)
                on^=i;
            else
                off^=i;
        }
        op.push_back((int)(on^xorpair));
        op.push_back((int)(off^xorpair));
        return op;
        
    }
};

int* singleNumber(int* nums, int numsSize, int* returnSize) {
    int i, *ret = calloc(*returnSize = 2, sizeof(int));
    for(i = 0; i < numsSize; ret[0] ^= nums[i++]);
    for(i = 0; i < numsSize; i++)
        if(nums[i] & ret[0] & -ret[0])
            ret[1] ^= nums[i];
    ret[0] ^= ret[1];
    return ret;
}