461. Hamming Distance

461. Hamming Distance (Easy)

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.

Related Topics:
Bit Manipulation

Similar Questions:

• Number of 1 Bits (Easy)
• Total Hamming Distance (Medium)

Solution 1.

// OJ: https://leetcode.com/problems/hamming-distance/
// Author: github.com/lzl124631x
// 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; 
    }
};

Solution 2.

// OJ: https://leetcode.com/problems/hamming-distance/
// Author: github.com/lzl124631x
// 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 &= z - 1) ++ans;
        return ans; 
    }
};