leetcode 191. Number of 1 Bits（bit中1的个数）

最新推荐文章于 2025-01-30 11:15:00 发布

原创最新推荐文章于 2025-01-30 11:15:00 发布 · 149 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode #算法

leetcode 专栏收录该内容

490 篇文章

订阅专栏

本文介绍了一种高效算法，用于计算32位整型数中1的比特位数量，即Hamming权重。该算法通过巧妙的位操作减少计算步骤，实现了快速统计。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

Note:

Note that in some languages, such as Java, there is no unsigned integer type. In this case, the input will be given as a signed integer type. It should not affect your implementation, as the integer’s internal binary representation is the same, whether it is signed or unsigned.
In Java, the compiler represents the signed integers using 2’s complement notation. Therefore, in Example 3, the input represents the signed integer. -3.

Example 1:

Input: n = 00000000000000000000000000001011
Output: 3
Explanation: The input binary string 00000000000000000000000000001011 has a total of three ‘1’ bits.
Example 2:

Input: n = 00000000000000000000000010000000
Output: 1
Explanation: The input binary string 00000000000000000000000010000000 has a total of one ‘1’ bit.

Hamming距离，input是32位整型，统计其中bit位为1的个数。

思路：
有个著名的并行统计bit为1的算法，
先是相邻2位，然后4位，8位，具体参考并行bit count算法

    public int hammingWeight(int n) {
        n = n - ((n >> 1) & 0x55555555);               
        n = (n & 0x33333333) + ((n >> 2) & 0x33333333); 
        int c = ((n + (n >> 4) & 0xF0F0F0F) * 0x1010101) >> 24;
        
        return c;
    }