---
title: 'LeetCode 461. Hamming Distance'
disqus: hackmd
---

# LeetCode 461. Hamming Distance

## 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

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.

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

## Constraints

0 <= x, y <= 2^31^ - 1

## Answer

Hamming Distance就是x要改變幾個位元才會變成y，ex:001 , 100 ,要改變最後一位變0然後第一位變1，總共兩位。此題可取兩者相異的位數，而取XOR時相異為1，所以只要取XOR算1的個數就知道要變幾個數。

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

## Link
https://leetcode.com/problems/hamming-distance/
###### tags: `Leetcode`