int hammingDistance(int x, int y) {
if((x^y)==0) return 0;
return (x^y)%2 + hammingDistance(x/2,y/2);
}
0ms C two line solution

The idea is good, it used bitwise operation methods to get the answer.
int hammingDistance(int x, int y) { if((x^y)==0) return 0; return (x^y) % 2 + hammingDistance(x/2, y/2); }
For example
14=1110
there are 3 1s in 14. You can use bitwise operation<<
to divide the number with 2 if the number is an even number else minus 1 and then continue divide 2 until the number is 0.