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.
My solution:
public class Solution {
public int hammingDistance(int x, int y) {
int temp = x ^ y;
int count = 0;
for(int i = 0; i < 32; i++) {
if ((temp & 1) == 1)
count++;
temp >>= 1;
}
return count;
}
}
大牛的解法：
Paste_Image.png
此解法利用了异或的重要性质，即 A ^ A = 0.
