# 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].

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.

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
0 1 1 2 1 2 2 3 1 2 2 3 2 3 3 4 1

numCount[i] = numCount[i-k] 其中k表示i只保留最高位的1时所代表的数。

dicuss中有一些更精妙的递推式：