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.
int hammingDistance(int x, int y)
给定两个32位的有符号型整数,当每一位不相同时,加上1,输出最后一共有多少位不相同。
代码如下:
int hammingDistance(int x, int y)
{
int xx,yy,count=0;
for(int i=0;i<32;i++)
{
xx=x&1;
yy=y&1;
if(xx!=yy)
count++;
x>>=1;
y>>=1;
}
return count;
}
代码浅显易懂,实测6ms,beats 72.58% of cpp submissions。
这道题在评论区还看到了令人耳目一新的做法,采用异或,当两位相同时输出0,当两位不同时输出1。不过这种做法的时间花费实际还是跟上述代码差不多的。
代码如下:
int hammingDistance(int x, int y)
{
int result=x^y,t1,count=0;
for(int i=0;i<32;i++)
{
t1=result&1;
if(t1==1)
count++;
result>>=1;
}
return count;
}