leetcode 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.

代码

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 * Return an array of size *returnSize.
 * Note: The returned array must be malloced, assume caller calls free().
 */
int* (int num, int* returnSize) {
    int* res = (int*)malloc((num+1)*sizeof(int));
    res[0] = 0;
    for (int i = 1; i <= num; i++ ) {
        res[i] = res[i>>1] + (i & 1);
    }
    *returnSize = num + 1; 
    return res;
}