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: 2Output: [0,1,1]

Example 2:

Input: 5Output: [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.

若為2的指數(shù)，則i&(i - 1)==0，如下圖所示，可以發(fā)現(xiàn)每個(gè)i值都是 i&(i-1) 對(duì)應(yīng)的值加1

i binary '1' i&(i-1)
0 0000 0
-----------------------
1 0001 1 0000
-----------------------
2 0010 1 0000
3 0011 2 0010
-----------------------
4 0100 1 0000
5 0101 2 0100
6 0110 2 0100
7 0111 3 0110
-----------------------
8 1000 1 0000
9 1001 2 1000
10 1010 2 1000
11 1011 3 1010
12 1100 2 1000
13 1101 3 1100
14 1110 3 1100
15 1111 4 1110

代碼如下：

class Solution {
    public int[] countBits(int num) {
        int[] result = new int[num + 1];
        result[0] = 0;
        for (int i = 1; i <= num; i++) {
            result[i] = result[i & (i - 1)] + 1;
        }
        return result;
    }
}