338. Counting Bits

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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

 1 /**
 2  * Return an array of size *returnSize.
 3  * Note: The returned array must be malloced, assume caller calls free().
 4  */
 5 int* countBits(int num, int* returnSize) {
 6     int *ary;
 7     int i;
 8     int k;
 9     ary = (int *)malloc((num + 1) * sizeof(int));
10     for(i = 0; i <= num; i++)
11     {
12         ary[i] = 0;
13         k = i;
14         while(k)
15         {
16             k &= (k-1);
17             ary[i]++;
18         }
19     }
20     *returnSize = num + 1;
21     return ary;
22 }

338. Counting Bits

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原文地址：http://www.cnblogs.com/boluo007/p/5495916.html