Formatted question description: https://leetcode.ca/all/461.html

# 461. Hamming Distance

Easy

## 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, 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.

## Solution

Compare corresponding bits of x and y directly. If the bits are different, add the Hamming distance by 1. Finally, return the Hamming distance.

Java

• class Solution {
public int hammingDistance(int x, int y) {
if (x == y)
return 0;
int distance = 0;
while (x > 0 || y > 0) {
int xBit = x & 0x1, yBit = y & 0x1;
if (xBit != yBit)
distance++;
x >>>= 1;
y >>>= 1;
}
return distance;
}
}

• // OJ: https://leetcode.com/problems/hamming-distance/
// Time: O(1)
// Space: O(1)
class Solution {
public:
int hammingDistance(int x, int y) {
int ans = 0;
for (int z = x ^ y; z; z >>= 1) ans += z & 1;
return ans;
}
};

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