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.

思路：
1.按位與運算（1&1=1, 1&0=0） ：res[n] = res[n & (n - 1)] + 1
2.移位運算：ans[n] = ans[n >> 1] + (n & 1)

註意：後面加逗號；

class Solution():
    
 
def countBits(self, num):
        """
        :type num: int
        :rtype: List[int]
        """
        res = [0]
        for i in range(1, num + 1):
            res += res[i & (i - 1)] + 1,
            # res += res[i >>1] + (i & 1),
        return res

338. Counting Bits (Medium)