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.

 

 

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网上查了，海明距离，其实就是两个数字进行异或运算，然后二进制结果中1的个数

我自己写了一个方法，比较笨，参考如下：

public class Solution {
    public int hammingDistance(int x, int y) {
        int hd=0;
        hd=x^y;
        String ids=Integer.toBinaryString (hd);
        int sum=0;
        for(int i=0;i<=ids.toCharArray().length-1;i++){
            if(ids.charAt(i)=='1'){
                sum++;
            }
        }
        
        return sum;
        
    }
}

 后来发现别人的更加先进：参考如下：

public class Solution {  
    public int hammingDistance(int x, int y) {  
        int res = x ^ y;  
        int count = 0;  
        for (int i = 0; i < 32; i++) {  
            if ((res & 1) != 0)  
                count++;  
            res >>= 1;  
        }  
        return count;  
    }  
}  

 别人是将异或运算和与运算，还有移位运算运用的很好