LeetCode338. Counting Bits还有更好的解法吗

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.

Credits:

大家都能想到的实现：

class Solution {
public:
    vector<int> countBits(int num) {
        vector<int> cb;
        for (int i = 0; i <= num; i++) {
            int count = 0;
            for (int c = 0; c < 32; c++) {
                if ((i >> c) & 0x1 == 1) {
                    count++;
                }
            }
            cb.push_back(count);
        }
        return cb;
    }
};

更优的一个实现：

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

还有更好的解法吗？？？？