Question:

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.
Hint:

You should make use of what you have produced already.
Divide the numbers in ranges like [2-3], [4-7], [8-15] and so on. And try to generate new range from previous.
Or does the odd/even status of the number help you in calculating the number of 1s?

Thinking:

We can divide the numbers into several groups and calculate the current elements based on the prvious ones. For example, in the group of [0-3] and [4-7], the later group’s elements have always one more 1 than the previous group.

Solution:

public int[] countBits(int num) {
    int[] res = new int[num+1];
    res[0] = 0;

    int flag = 1;
    for (int i = 1; i <= num; i++) {
        if (i / flag == 2)    flag *= 2;
        res[i] = res[i % flag] + 1;
    }
    return res;
}