找二进制表示中1的数量的方法

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本文介绍了一种计算32位无符号整数二进制表示中1的个数的方法，即汉明重量。文章提供了两种算法实现：一种通过逐位左移并进行按位与操作来计数；另一种更高效的算法利用“n &= n-1”技巧删除最右侧的1，直至n变为0，其时间复杂度依赖于1的个数。

题目来源:

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题目描述:

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.

我的解决方案:

class Solution {
public:
    int hammingWeight(uint32_t n) {
        uint32_t tmp=1;
        int ret=0;
        while(tmp)
        {
            if(tmp&n)
              ret++;
            tmp=tmp<<1;
        }
        return ret;
    }
};

很简单也很好想的暴力算法,把1挨个移位,然后相与

别人的算法:

int hammingWeight(uint32_t n)
{
    int res = 0;
    while(n)
    {
        n &= n - 1;
        ++ res;
    }
    return res;
}

源作者

makuiyu

"n &= n - 1" is used to delete the right "1" of n. For example:

if n = 5 (101), then n-1 = 100, so n & (n-1) = 100, the right "1" is deleted;
if n = 6,(110), then n-1 = 101, so n & (n-1) = 100, the right "1" is also deleted;
and so on...

So time complexity is O(m), and m is the count of 1's, also m is less than or equal to 32.