Minimum Bit Flips to Convert Number

给两个数字, 求两个数字中不相同的bit的个数.

先xor, 然后counting bit.

class Solution {
    public int minBitFlips(int start, int goal) {
        int x = start ^ goal;
        int res = 0;
        for (res = 0; x != 0; x >>= 1)
        {
          res += x & 1;
        }
        return res;
    }
}