版权声明:本文为博主原创文章,未经博主允许不得转载。 https://blog.csdn.net/u011391629/article/details/86530111
Question
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.
Algorithm
可以一位一位的去比,我这里用异或表示,然后统计二进制中1的个数
Accepted Code
class Solution {
public:
int hammingDistance(int x, int y) {
int t = x^y;
int res = 0;
while(t){
if(t & 0x01 == 1)
res++;
t = t >> 1;
}
return res;
}
};
换一种写法:
class Solution {
public:
int hammingDistance(int x, int y) {
int t = x^y;
int res = 0;
while(t){
res++;
t = t & (t-1);
}
return res;
}
};