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

461. Hamming Distance

Level

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
    
    

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