338. 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 1:

Input: 2
Output: [0,1,1]

Example 2:

Input: 5
Output: [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.

Easy DP

public int[] countBits(int num) {
    int[] dp = new int[num+1];
    for(int i = 0; i <= num; i++) {
        if((i & 1) == 1) {
            dp[i] = dp[i-1]+1;
        } else {
            dp[i] = dp[i >> 1];
        }
    }
    return dp;
}