1 解題思想

這道題就是求兩個數的海明距離，而海明距離也就是兩個數對應的二進位制bit位裡不同的位數

所以這道題直接異或就可以，異或時，二者相同為0，不同為1，異或完後統計為1的位數就好

2 原題

he 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.

3 AC解

public class Solution {
    public int hammingDistance(int x, int y) {
        return Integer.bitCount(x ^ y);
    }
}