Problem：
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. Code1:
public class Solution {
public int hammingDistance(int x, int y) {
return Integer.bitCount(x ^ y);
}
}
Code2;
public class Solution {
public int hammingDistance(int x, int y) {
int n = (x ^ y);
int count = 0;
while(n > 0){
n &= n-1;//除去最右边的1
count++;//得到n里面1的个数
}
return count;
}
}
