https://leetcode.com/problems/hamming-distance/description/
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.
思路:先将x与y异或,得到temp,其二进制表示中的"1"的个数即为海明距离.将temp向右移i位(i从0到30)然后和"1"与操作,即检测最右一位是否为1,是的话count++.
class Solution {
public:
int hammingDistance(int x, int y) {
int count = 0;
int temp = x ^ y;
for(int i = 0; i <= 30; i++){
count += (temp >> i) & 1;
}
return count;
}
};
