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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

题意:给定两个数,求出其汉明距离

题解:将两个数异或,那么二进制位上相同的都变成0,不同变成1,统计一下有多少个1即可

 
class Solution {
public:
//6ms
int hammingDistance(int x, int y) {
int res = 0;
while (x && y){
if ((x & 1) != (y & 1)) res++;
x >>= 1;
y >>= 1;
}
while (x){
if (x & 1) res++;
x >>= 1;
}
while (y){
if (y & 1) res++;
y >>= 1;
}
return res;
} //3ms
int hammingDistance(int x,int y){
x ^= y;
y = 0;
while (x){
y += (x & 1);
x >>= 1;
}
return y;
}
};

  

 

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