题目描述

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 < 2^31.

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.

分析

题目可以理解为：求两个数有多少位是不一样的。那么步骤可以为：

1. 将两个数异或，结果记为 val
2. 计算 val 有多少个 1

代码

public int hammingDistance(int x, int y) {
    int val = x ^ y;
    int result = 0;
    while (val != 0) {
        if ((val & 1) == 1) {
            result++;
        }
        val >>= 1;
    }
    return result;  
}