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:

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?

Space complexity should be O(n).

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

公众号每日一题有详细讲解分析：

P 题的难点在于构建最优子结构。只要构造了最优子结构，一般就比较容易解题了。构造最优子结构的方法有两种：倒推发现 n 与 0 到 n - 1 的关系；使用 DFS 和 递归 方法，从 0 到 n 发现规律。

今天这道题的解法试图通过观察其特点发现规律。这也是一般性的思考问题的方法：通过画图，将抽象的问题具体化，通过具体的例子来发现规律，最后推广到 n 的规模。

class Solution {
public:
    vector<int> countBits(int num) {
        vector<int> res(num + 1, 0);
        res[1] = 1;
        for ( int i = 0; i <= num; i++)
        {
            if ( i % 2 == 0)
            {
                res[i] = res[i/2];
            }
            else
            {
                res[i] = res[i/2] + 1;
            }
        }
        return res;
    }
};

class Solution {
public:
    vector<int> countBits(int num) {
        vector<int> res(num + 1, 0);
        for ( int i = 1; i <= num; i++)
        {
            res[i] = res[i >> 1] + (i & 1);
        }
        return res;
    }
};

LeetCode 338. Counting Bits