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:
Fornum = 5you 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.

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

Thoughts:

1. f[i] = f[i/2] (right shift by 1) + f[i%2] with f[0] = 0 and f[1] =1;

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

        return f;
    }
};