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 1:

Input: 2
Output: [0,1,1]

Example 2:

Input: 5
Output: [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:
    def countBits(self, num):
        """
        :type num: int
        :rtype: List[int]
        """
        res = []
        for i in range(num+1):
            res.append(bin(i).count('1'))
        return res
原文地址:https://www.cnblogs.com/bernieloveslife/p/9798382.html