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.
My code:
class Solution(object):
def countBits(self, num):
"""
:type num: int
:rtype: List[int]
"""
result=[]
for k in range(num+1):
result.append(self.calcu_bit(k))
return result
def calcu_bit(self,k):
if k==0:
return 0
count=0
while k>0:
d=k>>1
if d==0:
count+=1
break
if k%2==1:
count+=1
k=d
return count