[LeetCode]461. Hamming Distance

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.

汉明距离：两个数用二进制表示时，两数对应位不同的个数。可将两数异或，再算得出结果的各个位上的1的个数。

如十进制除以10，求二进制数各个位上1的个数，可以将其一直除以2，将其余数个数相加。

1版：

class Solution {
public:
int hammingDistance(int x, int y) {

int temp;
int i = 0;//计数
temp = x^y;
while (temp / 2 != 0)
{
if (temp % 2 == 1)
i++;
temp = temp / 2;
}
if (temp == 1)
i++;
return i;
}
};

2.看到 x&(x-1) 可以消去x最右位的1

2版

class Solution {
public:
int hammingDistance(int x, int y) {
int i = 0;
int temp = x^y;
while (temp)
{
temp &= (temp - 1);
i++;
}
return i;
}
};



3版 简化1

class Solution {
public:
int hammingDistance(int x, int y) {
if ((x^y) == 0)
return 0;
return (x^y) % 2 + hammingDistance(x / 2, y / 2);
}
};