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.
这种题真的会有各种技巧，先上个自己写的：
1 class Solution { 2 public: 3 int hammingDistance(int x, int y) 4 { 5 int z = x ^ y; 6 int res = 0; 7 while (z != 0) 8 { 9 int mod = z % 2; 10 res += mod; 11 z >>= 1; 12 } 13 return res; 14 } 15 };
贴一个优化过的：
1 class Solution { 2 public: 3 int hammingDistance(int x, int y) { 4 int z = x^y; 5 int count = 0; 6 while(z){ 7 count += z&1; 8 z >>= 1; 9 } 10 return count; 11 } 12 };