leetcode 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.
假设最右的1位于m位，

(n-1)是把m位左边的数取反，右边的数保留。

(n-1)&n 就可以将m位左边的数保留，右边的数抹掉(取反后相与=0)。重复此操作，直到m=0即可遍历出二进制表达式中所有的1.

/**
* Return an array of size *returnSize.
* Note: The returned array must be malloced, assume caller calls free().
*/

int NumOf1(int n){
int count =0 ;

while(n)
{
count++;
n = (n - 1) & n;
}

return count;
}

int* countBits(int num, int* returnSize) {
*returnSize = num+1;
int *ptr = (int *)malloc((sizeof(int))*(num+1));
for(int i=0;i<=num;i++)
{
ptr[i] = NumOf1(i);
}

return ptr;
}