[LeetCode] 461. Hamming Distance 汉明距离

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.

 

两个数字之间的汉明距离是其二进制数对应位不同的个数，两个数异或，把为1的数量累加起来就是汉明距离。

Java:

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

Python:

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

Python:

class Solution(object):
    def hammingDistance(self, x, y):
        """
        :type x: int
        :type y: int
        :rtype: int
        """
        return bin(x ^ y).count('1')　

C++:

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

C++:　　

class Solution {
public:
    int hammingDistance(int x, int y) {
        int res = 0, exc = x ^ y;
        while (exc) {
            ++res;
            exc &= (exc - 1);
        }
        return res;
    }
};

C++:

class Solution {
public:
    int hammingDistance(int x, int y) {
        if ((x ^ y) == 0) return 0;
        return (x ^ y) % 2 + hammingDistance(x / 2, y / 2);
    }
};

　　

类似题目：　　

[LeetCode] 191. Number of 1 Bits 二进制数1的个数

 

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