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.
两个字码中不同位值的数目称为汉明距离(Hamming distance)，由此可得此题用XOR求解。
class Solution { public: int hammingDistance(int x, int y) { int res = x ^ y; int count = 0; while (res) { if (res & 1) count++; res >>= 1; } return count; } }; // 3 ms
内容来源于网络如有侵权请私信删除
- 还没有人评论，欢迎说说您的想法！
客服
