[LC]461. Hamming Distance

一、问题描述

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.

二、我的思路

首先想到用位运算。

先把x和y异或（^），记作xor，里面是1的表示x与y不同的位，是0的表示相同的位。然后统计这个数字的二进制里面有多少个1就行了～

统计的时候每次把xor和1做与（&）运算，加到count里。因为规定了它们的最大值，所以循环最多32次，统计32位的情况。

代码：

class Solution {
    public int hammingDistance(int x, int y) {
        int xor = x ^ y;
        int count = 0;
        for(int i = 0; i < 32; i ++){
            count += xor & 1;
            xor = xor >> 1;
        }
        return count;
    }
}

三、别人的淫奇技巧

1. 用我没见过的函数……

public class Solution {
    public int hammingDistance(int x, int y) {
        return Integer.bitCount(x ^ y);
    }
}

2. 方法类似，短小精悍。。

public int hammingDistance(int x, int y) {
    int xor = x ^ y, count = 0;
    for (int i=0;i<32;i++) count += (xor >> i) & 1;
    return count;
}

3. 这个有点高端的样子

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

看不懂啊，这是什么…引用大神的解答：

@jonvasquez1 Considering this example. Let's say n = 10101, and dist = 0 asccording to above code of @pengr7.

• Before we go into the while loop. n = 10101, dist = 0
• Loop 1.  dist = 1, n =10101 & 10100 = 10100
• Loop 2.  dist = 2, n =10100 & 10011 = 10000
• Loop 3.  dist = 3, n =10000 & (0)1111 = 0
• Loop ends. dist = 3

The change of n :

• 10101
• 10100
• 10000
• 00000

Conclusion: n & (n-1) converts the right most 1 to 0 .
This is the key idea solving the problem. By counting how many times we can perform this operation, we know how many 1 exists in the binary representation of n

大概意思就是，n&(n-1)会把二进制最右边的1变成0。然后有几个1就能变成几个不为0的值。简直6666～差距吖

四、知识点

位运算。

参考博客：http://blog.youkuaiyun.com/xiaochunyong/article/details/7748713

五、碎碎念

对Java的位运算符不是特别熟悉。。继续加油～