[LeetCode] Hamming Distance（二进制中有多少个1）+ Number Complement（补码）

啊，又是好久没来了！

最近又开始刷leetcode，发现题目多了不少。还是从基础开始做吧。

这里是两道关于位运算的题。

1. Hamming Distance

题目链接在此

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.

就是数一下两个二进制数之间有位数有区别。很简单，先异或，再数有多少个1。

输多少个1那一步，我写的这篇博客有。

这里是我在不记得大神的做法的情况下，自己写的（3ms）：

class Solution {
public:
int hammingDistance(int x, int y) {
int a = x ^ y;

int count = 0;

while (a != 0) {
count += (a & 1);
a >>= 1;
}

return count;
}
};

[b]2. Number
Complement[/b]

题目链接在此

Given a positive integer, output its complement number. The complement strategy is to flip the bits of its binary representation.

Note:

The given integer is guaranteed to fit within the range of a 32-bit signed integer.
You could assume no leading zero bit in the integer’s binary representation.

Example 1:

Input: 5
Output: 2
Explanation: The binary representation of 5 is 101 (no leading zero bits), and its complement is 010. So you need to output 2.

Example 2:

Input: 1
Output: 0
Explanation: The binary representation of 1 is 1 (no leading zero bits), and its complement is 0. So you need to output 0.

补码定义为二进制每位都翻转后得到的数字。

这是我写的，就是先算出这个数的二进制有多少位，然后用这么多位的1跟他做异或。比如2的二进制是10，有两位， 就拿11和10做异或，得到01。显然比较慢：

class Solution {
public:
int findComplement(int num) {
int n = num;
int bits = 0;
while(n) {
bits++;
n >>= 1;
}
return num ^ int(pow(2.0, bits) - 1);
}
};

这是discuss里的大神的做法：

class Solution {
public:
int findComplement(int num) {
unsigned mask = ~0;
while (num & mask) mask <<= 1;
return ~mask & ~num;
}
};

例子：

num          = 00000101
mask         = 11111000
~mask & ~num = 00000010