Example 1:
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Input: 2
Output: [0,1,1]
Example 2:
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Input: 5
Output: [0,1,1,2,1,2]
Follow up:
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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.

class Solution {
public:
    vector<int> countBits(int num) {
        int s= log(num)/log(2);
        if(pow(2,s) <num) s+=1;
        int N= 1;
        int count =1;
        vector<int> res(num+1, 0);
       for(int i=1;i<=num;i++)
       {
           int count =0;
           for(int j=0;j<=s;j++)
           if(i & (1<<j)) count++;
           res[i] = count;
       }
        return res;
    }
};

class Solution {
vector <int > counts={0,1,1,2,1};
public:
    vector<int> countBits(int num) {
      if(num <counts.size()) 
      {
          vector<int > res(counts.begin(), counts.begin()+num+1);
          return res;
      }
      for(int i=counts.size(); i<=num;i++)
      counts.push_back( counts[i/2] + i%2);
       return counts;
    }
};