461. Hamming Distance(easy)

Easy

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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.

 

C++:

/*
 * @Autor: SourDumplings
 * @Date: 2019-09-26 17:03:36
 * @Link: https://github.com/SourDumplings/
 * @Email: [email protected]
 * @Description: https://leetcode.com/problems/hamming-distance/
 */

class Solution
{
public:
    int hammingDistance(int x, int y)
    {
        int r = x ^ y;
        int res = 0;
        for (int i = 0; i != 32; ++i)
        {
            if (r & 1 == 1)
            {
                ++res;
            }
            r >>= 1;
        }
        return res;
    }
};

 Java:

/*
 * @Autor: SourDumplings
 * @Date: 2019-09-26 17:10:24
 * @Link: https://github.com/SourDumplings/
 * @Email: [email protected]
 * @Description: https://leetcode.com/problems/hamming-distance/
 */

class Solution
{
    public int hammingDistance(int x, int y)
    {
        int r = x ^ y;
        int res = 0;
        for (int i = 0; i != 32; ++i)
        {
            if ((r & 1) == 1)
            {
                ++res;
            }
            r >>= 1;
        }
        return res;
    }
}