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