题目描述

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(nsizeof(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.

Credits:

Special thanks to @ syedee for adding this problem and creating all test cases.

代码实现

class Solution:
    def countBits(self, num):
        """
        :type num: int
        :rtype: List[int]
        """
        ans=[]
        for i in range(num+1):
            ans.append(int(bin(i).count('1')))
        return ans