The Hamming distance between two integers is the number of positions at which the corresponding bits are different.

Given two integers x and y, calculate the Hamming distance.

Note :

0 <= x, y <= 2^31.

Example :

Input: x = 1, y = 4  
  
Output: 2  
  
Explanation:  
1   (0 0 0 1)  
4   (0 1 0 0)  
       ↑   ↑  
  
The above arrows point to positions where the corresponding bits are different.  

solution

time complexity : O(1)
space complexity : O(1)

class Solution {
public:
    int hammingDistance(int x, int y) {
        return __builtin_popcount(x ^ y);
    }
};

此題在問，計算兩數在二進位表示時，有幾個位元不同。
做法是將兩數取xor後，即可讓相同的bit為0，不同的bit為1。
再將xor後的數，一個bit一個bit計數，即為答案。