[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:

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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:

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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:

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

C++:

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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++:　　

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

C++:

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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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