Leetcode16: Number of 1 Bits

Write a function that takes an unsigned integer and returns the number of ’1‘ bits it has (also known as the Hamming
weight).

For example, the 32-bit integer ’11‘ has binary representation 00000000000000000000000000001011,
so the function should return 3.

class Solution {
public:
    int hammingWeight(uint32_t n) {
        int k = 0;
        while(n)
        {
            k += (n & 0x1) > 0 ? 1 : 0; //(n % 2) > 0 ? 1 : 0;
            n >>= 1;                    //n /= 2;
        }
        return k;
    }
};

这是比较容易想到的解决办法，每次判断最后一位是否是1。但是这样32位的数最坏的情况要比较32次。有没有更简单的方法呢？

下面这种方法更为简单。假设n＝
1111000111000 那 n-1 = 1111000110111,
(n-1) & n = 1111000110000，刚好把最后一个1给干掉了。也就是说，
(n-1)&n 刚好会从最后一位开始，每次会干掉一个1.这样速度就比上面哪种方法快了。有几个1，就执行几次。（学渣表示震惊！= =）

class Solution {
public:
    int hammingWeight(uint32_t n) {
        int k = 0;
        while(n != 0)
        {
            n = n & (n-1);
            k++;
        }
        return k;
    }
};