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.
题意:求两个数x、y的二进制下各位不同的总数。
用位运算^异或,每个位相同为0,不同则为1。
class Solution {
public:
int hammingDistance(int x, int y) {
int res=0;
int n = x^y;
while(n)
{
res++;
n=n&(n-1);
}
return res;
}
};
/*直接模拟*/
class Solution {
public:
int hammingDistance(int x, int y) {
int res=0;
int cnt_x=0,cnt_y=0;
while(x&&y)
{
if(y%2!=x%2)res++;
y/=2,x/=2;
}
while(x)
{
if(x%2) res++;
x/=2;
}
while(y)
{
if(y%2) res++;
y/=2;
}
return res;
}
};