计数位中1的数量-优快云博客

本文介绍了一种高效计算从0到给定整数num范围内每个数字二进制表示中1的数量的方法。通过利用2^n的特性，文章提供了一个线性时间复杂度O(n)的C++解决方案。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

338.Counting Bits

Description：

Given a non-negative integer number num. For every numbers i in the range 0 ≤ i ≤ num calculate the number of 1's in their binary representation and return them as an array.

Example: For num = 5 you should return [0, 1, 1, 2, 1, 2].

Follow up:

l It is very easy to come up with a solution with run time O(n*sizeof(integer)). But can you do it in linear time O(n) /possibly in a single pass?

l Space complexity should be O(n).

l Can you do it like a boss? Do it without using any builtin function like __builtin_popcount in c++ or in any other language.

分析：

注意到2^n二进制表示只有一位为1, 以这些数字作为分割可以“平移”子问题的结果。所以迭代递推即可O(n)求解。

My C++ code：

class Solution {
public:
    vector<int> countBits(int num) {
        vector<int> m(num+1, 0);
        for(int i = 1; i <= num; i = i * 2)
        {
            m[i] = 1;
        }
        for(int i = 1; pow(2, i) <= num; ++i)
        {
            for(int j = pow(2, i) + 1; j < pow(2, i+1) && j <= num; ++j)
            {
                m[j] = 1 + m[j-pow(2, i)];
            }
        }
        return m;
    }
};