Problem

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.

Remark

这道题的大体意思就是给出一个N，求出从0到N中所有数字，转成2进制时1的个数。

Solution

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class Solution:
    def countBits(self, num):
        """
        :type num: int
        :rtype: List[int]
        """
        if num == 0:
            return [0]
        if num == 1:
            return [0, 1]

        count = [0, 1]
        index = 0
        flag = 2

        for i in range(2, num+1):
            if i == flag:
                flag *= 2
                index = 0

            count.append(count[index]+1)
            index += 1

        return count

我还没有理解这种解法。

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class Solution2:
    def countBits(self, num):
        ret = [0]
        for i in range(1, num + 1):
            ret.append(ret[i&(i-1)]+1)
        return ret