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:
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.

#include <iostream>
#include <vector>
using namespace std;
class Solution {
public:
    vector<int> countBits(int num) {
        vector<int> r;
        r.push_back(0);
        for(int i=1;i<=num;i++)
        {
        int re = r[i >> 1] + (i & 1);//根据规律，若num的二进制表示最后一位为0，则它的1个数等于它除以2后得到的数中的1的个数，若最后一位为1，则需要再加1
r.push_back(re); 
}
        return r;
    }
};