Counting Bits

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:

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

Credits:
Special thanks to @ syedee for adding this problem and creating all test cases.

public class Solution {
    public int[] countBits(int num) {
        int[] res = new int[num+1];
        int pow = 1;
        res[0] = 0;
        for(int i=1,t=0;i<=num;i++,t++)
        {
            if(i==pow)
            {
                pow = pow * 2;
                t = 0;
            }
            res[i] = res[t] + 1;
            System.out.println(res[i]);
        }
        return res;
        
    }
}

reference: https://discuss.leetcode.com/topic/41785/simple-java-o-n-solution-using-two-pointers/3