Hamming Distance(汉明距离)

Introduction

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.

解法一：

两个数字之间的汉明距离就是其二进制数对应位不同的个数。按位分别取出两个数对应位上的数并异或，满足条件则累加

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

解法二：

直接将两个数字异或起来，遍历异或结果的每一位并统计个数

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

解法三：

最高效的解法

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

解法四：

递归终止的条件是当两个数异或为0时，表明此时两个数完全相同，我们返回0，否则我们返回异或和对2取余加上对x/2和y/2调用递归的结果。异或和对2取余相当于检查最低位是否相同，而对x/2和y/2调用递归相当于将x和y分别向右移动一位，这样每一位都可以比较到。

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

参考资料