leetcode刷题(python)－－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

class Solution(object):
    def countBits(self, num):
        """
        :type num: int
        :rtype: List[int]
        """
        res = [0]
        while len(res) <= num:
            res += [x +1 for x in res]
        return res[:(num+1)]

注意这里利用了十进制转为二进制时的一个规律， 每次扩充二进制位数时，二进制数中1的数目总是在之前的基础上依次加1。

比如如果num = 15

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

其中1 的数目依次更新： 

0 

0    1

0  1    1  2

0  1  1  2    1  2  2  3

0  1  1  2  1  2  2  3    1  2  2  3  2  3  3  4