461. Hamming Distance

题目描述【Leetcode】

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.

这道题是比较两个数的二进制形式的不同位的个数。
方法一：
用数组来将每个数二进制的每一位，再进行比较，这个比较麻烦：

class Solution {public:    int hammingDistance(int x, int y) {        vector<int>q1;        vector<int>q2;        int count = 0;        while(x  != 0){            int m = x%2;            x = x/2;            q1.push_back(m);        }        while(y != 0){            int m = y%2;            y = y/2;            q2.push_back(m);        }        while(q1.size() > q2.size()){                q2.push_back(0);        }        while(q1.size() < q2.size()){                q1.push_back(0);        }        for(int i = 0; i < q1.size(); i++){            if(q1[i] != q2[i]) count++;        }        return count;    }};

方法二：利用bitset来做：

class Solution {public:    int hammingDistance(int x, int y) {        bitset<64>b1(x);        bitset<64>b2(y);        int count = 0;        for(int i = 0; i < b1.size(); i++){          if ((b1.test(i) && !b2.test(i)) || (!b1.test(i) && b2.test(i)))               count++;        }        return count;    }};

方法三：用按位“异或”运算符和按位“与”赋值运算符来做，这种办法运行起来更快（运算符参考资料）

class Solution {public:    int hammingDistance(int x, int y) {    int n = x^y;     // n为x与y的异或    int count = 0;    while(n) {        if((n >> 1) << 1 != n) // 判断n右移一位再左移一位是否会变          count++;        n >>= 1;    }    return count;}};