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.
代码：
/**
* Return an array of size *returnSize.
* Note: The returned array must be malloced, assume caller calls free().
*/
int *countBits(int n,int *returnSize)
{
int *p = (int *) malloc(sizeof(int)*(n+1));
*returnSize = n+1;
int tmp = 2,i = 2;
int *q = p;
*q++ = 0;
*q++ = 1;
while(i<=n)
{
*q = *(q-tmp)+1;
q++;
i++;
if(i%tmp == 0)
tmp = tmp<<1;
}
return p;
}