[leetcode]461. Hamming Distance

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

 <
231.

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.

思路：

(1)先求x^y的结果res。

(2)再依次求res的每一位与1进行与操作的结果，若不为0，则Hamming Distance加一。

(3)每判断完一位，res右移一位继续判断下一位
public class Solution {
public int hammingDistance(int x, int y) {
int count = 0;
int res = x^y;
while(res>0) {
if((res&1)!=0) {
count++;
}
res>>=1;
}
return count;
}
}