LeetCode 461 - Hamming Distance

http://bookshadow.com/weblog/2016/12/18/leetcode-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.

https://leetcode.com/problems/hamming-distance/discuss/94698/Java-1-Line-Solution-%3AD

What does come to your mind first when you see this sentence "corresponding bits are different"? Yes, XOR! Also, do not forget there is a decent function Java provided: Integer.bitCount() ~~~

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

https://leetcode.com/problems/hamming-distance/discuss/94713/Java-solution

    public int hammingDistance(int x, int y) {
        int xor = x ^ y, count = 0;
        
        while (xor != 0) {
            xor &= (xor - 1);
            count++;
        }
        return count;
    }

https://leetcode.com/problems/hamming-distance/discuss/94705/My-C%2B%2B-solution-using-bit-manipulation

    int hammingDistance(int x, int y) {
        int dist = 0, n = x ^ y;
        while (n) {
            ++dist;
            n &= n - 1;
        }
        return dist;
    }

https://leetcode.com/problems/hamming-distance/discuss/94693/Java-3-Line-Solution

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

http://blog.csdn.net/zhouziyu2011/article/details/53726977

1.     public int hammingDistance(int x, int y) {  
2.         int res = x ^ y;  
3.         int count = 0;  
4.         for (int i = 0; i < 32; i++) {  
5.             if ((res & 1) != 0)  
6.                 count++;  
7.             res >>= 1;  
8.         }  
9.         return count;  
10.     }