README_EN.md

461. Hamming Distance

中文文档

Description

The Hamming distance between two integers is the number of positions at which the corresponding bits are different.

Given two integers x and y, return the Hamming distance between them.

 

Example 1:

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.

Example 2:

Input: x = 3, y = 1
Output: 1

 

Constraints:

• 0 <= x, y <= 231 - 1

Solutions

Use xor operation to find different bits.

• 0 ^ 0 = 0
• 1 ^ 1 = 0
• 0 ^ 1 = 1
• 1 ^ 0 = 1

Python3

class Solution:
    def hammingDistance(self, x: int, y: int) -> int:
        num, count = x ^ y, 0
        while num != 0:
            num &= num - 1
            count += 1
        return count

Java

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

Or use the library function Integer.bitCount()

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

JavaScript

/**
 * @param {number} x
 * @param {number} y
 * @return {number}
 */
var hammingDistance = function (x, y) {
    let distance = x ^ y;
    let count = 0;
    while (distance != 0) {
        count++;
        distance &= distance - 1;
    }
    return count;
};

C++

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

Go

func hammingDistance(x int, y int) int {
	x ^= y
	count := 0
	for x != 0 {
		count++
		x &= (x - 1)
	}
	return count
}

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