Counting Number of 1 Bits in an Integer

Guess Version #1: LC191. Number of 1 Bits

Write a function that takes an unsigned integer and returns the number of '1' bits it has (also known as the Hamming weight).

For example, the 32-bit integer ’11' has binary representation 00000000000000000000000000001011, so the function should return 3.

Solution #1: O(n) time & O(1) Space

class Solution(object):
    def hammingWeight(self, n):
        """
        :type n: int
        :rtype: int
        """
        cont = 0
        for i in bin(n):
            if i == '1':
                cont += 1
        return cont

Solution #2: n & (n-1) trick: O(n) time & O(1) Space

class Solution(object):
    def hammingWeight(self, n):
        """
        :type n: int
        :rtype: int
        """
        cont = 0
        while n:
            n &= n-1
            cont += 1
        return cont

Guess Version #2: LC338. 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.

Hint:

You should make use of what you have produced already.
Divide the numbers in ranges like [2-3], [4-7], [8-15] and so on. And try to generate new range from previous.
Or does the odd/even status of the number help you in calculating the number of 1s?

Solution:

 class Solution(object):
    def countBits(self, num):
        """
        :type num: int
        :rtype: List[int]
        """
        if num < 2:
            return [0, 1] if num == 1 else [0]
        total = [0, 1]
        for i in xrange(2, num+1):
            if i % 2 == 0:
                total.append(total[i/2])
            else:
                total.append(total[i/2] + 1)
        return total

Counting Number of 1 Bits in an Integer