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:

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.
Method1:
 https://discuss.leetcode.com/topic/53702/28ms-c-solution-o-n-time-o-n-space

def countBits0(self, num):
"""
:type num: int
:rtype: List[int]
"""
result = [ 0 for i in range(num+1)]
result[0]= 0
i=1
while i<=num :
cnt=i
cnt&=(cnt-1)
result[i] = result[cnt]+1
i +=1
return result

Method2: https://discuss.leetcode.com/topic/44008/5-lines-of-python-o-n-time-simple-no-bitwise-operation/2

def countBits(self, num):
r = [0]*(num+1)
a = 1
for i in range(1, num+1):
r[i]=r[i-a]+1
if i == 2*a - 1: a *= 2
return r

r.append is slower

Method3: https://discuss.leetcode.com/topic/40162/three-line-java-solution/25

def countBits(self, num):
result = [0]*(num+1)
for i in range (1,num+1):
result[i]= result[i>>1]+(i&1)
return result

Method4: https://discuss.leetcode.com/topic/41785/simple-java-o-n-solution-using-two-pointers

def countBits(self,num):
result = [0]*(num+1)
result[0] = 0
power =1
for i in range(1,num+1):
if (i==power):
power *=2
t=0
result[i]=result[t]+1
t +=1
return result