[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 = 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.

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[
4000
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。