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 timeO(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.

计算0 ≤ i ≤ num内所有数字二进制的“1“”的个数

i   二进制 个数 i&(i-1)

0  0000    0      0000

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

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