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.

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.Show
More Hint 

Credits:

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

求0~num的数，每个数的二进制表示中分别有多少个1。

已经给提示了，2的幂次数1的个数只有1个，每次找出当前的数和哪个2的幂次最接近，把这个作为基数（其实就是1），

然后当前搜索数=（当前数-离当前数最近的2的幂次），这个当前搜索数一定在前面求出来了，直接加上就好。

public static int[] countBits(int num)
{
int[] arr=new int[num+1];
if(num<2)
{
for(int i=0;i<=num;i++)
arr[i]=i;
return arr;
}

arr[0]=0;
arr[1]=1;
arr[2]=1;

double lg2=Math.log10(2);

for(int i=3;i<=num;i++)
{
int power=(int) (Math.log10(i)/lg2);
int base=(int) Math.pow(2, power);
arr[i]=arr[base]+(i-base==0?1:arr[i-base]);
}

return arr;
}