[LeetCode] 461. Hamming Distance

Problem:

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 = 4Output: 2Explanation:1   (0 0 0 1)4   (0 1 0 0)       ↑   ↑The above arrows point to positions where the corresponding bits are different.

Solution 1:

（傻瓜法：把两个十进制数转化为二进制串，为了比较各个位是否相同，需要对较短的字符串补"0"）

class Solution {public:    int hammingDistance(int x, int y) {        string num1 = convertToBinary(x);        string num2 = convertToBinary(y);        return countDifferentBits(num1, num2);    }        string convertToBinary(int num) {        string str = "";        for (int i = num; i > 0; i /= 2) {            str += ((i % 2) ? '1' : '0');        }        reverse(str.begin(), str.end());        return str;    }    void modifyString(string& s, int n) {    string str = "";    for (int i = 0; i < n; i++) {    str += "0";    }    for (int i = 0; i < s.length(); i++) {    str += s[i];    }    s = str;    }        int countDifferentBits(string num1, string num2) {        int count = 0;        int len1 = num1.length();        int len2 = num2.length();        if (len1 < len2) modifyString(num1, len2-len1);        if (len2 < len1) modifyString(num2, len1-len2);        for (int i = 0; i < num1.length(); i++) {            if (num1[i] != num2[i]) {                count++;            }        }        return count;    }};

Solution 2:

（利用异或操作：两个数作异或，相同的位则为0，不同则为1，得到的结果转化为二进制串，计算字符"1"的数目，即为所求

的海明距离）

class Solution {public:    int hammingDistance(int x, int y) {        int z = x ^ y;        string str = convertToBinary(z);        return countDifferentBits(str);    }        // 把十进制数转化成二进制字符串    string convertToBinary(int num) {        string str = "";        for (int i = num; i > 0; i /= 2) {            str += ((i % 2) ? '1' : '0');        }        reverse(str.begin(), str.end());        return str;    }    // 计算海明距离    int countDifferentBits(string num) {        int count = 0;        for (int i = 0; i < num.length(); i++) {            if (num[i] == '1') {                count++;            }        }        return count;    }};

Solution 3 :

（解法出自：https://discuss.leetcode.com/topic/72236/my-c-solution-using-bit-manipulation）

（利用位运算：异或 + 与；首先求得两个数的异或，再通过与运算来计算海明距离）

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

Note：这里理解n &= n-1是一个难点，这个式子的作用是把n最右边的1转化为0，以此来统计异或结果中1的数目，即海明距离。