461. Hamming Distance

题目：

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.

链接：https://leetcode.com/problems/hamming-distance/#/description

3/27/2017

public class Solution {
    public int hammingDistance(int x, int y) {
        int z = x ^ y;
        int c = 1, count = 0;
        while (z != 0) {
            if ((z & c) != 0) count++;
            z >>= 1;
        }
        return count;
    }
}

其他人作法：

1. 使用自带函数

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

更多讨论：https://discuss.leetcode.com/category/590/hamming-distance