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.

解法一

思路

汉明距离，即两个数的二进制表示中对应位上数字（0／1）不相同的个数。如01101和01011的汉明距离即为2，因其仅第1和第2位（从右至左，0开始）不相同。

很容易想到这题应该用异或，即

01101 ^ 01011 = 00110

现在数00110中1的个数就好了。

怎么数1的个数？这里有个简单的方法：

num = 0b00110
count = 0
while num:
    num &= num - 1
    count += 1
print(count)

每次num &= num - 1，num二进制值的最低位1都会变成0，即每次去掉一个1，直到其值为0，举个例子看看：

00110 & 00101 = 00100
00100 & 00011 = 00000

这样汉明距离就算出来啦～

Python

class Solution(object):
    def hammingDistance(self, x, y):
        """
        :type x: int
        :type y: int
        :rtype: int
        """
        count = 0
        xor = x ^ y
        while xor != 0:
            xor &= (xor - 1)
            count += 1
        return count

Java

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

C++

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