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].

From:

1. Naive Solution

We can simply count bits for each number like the following:

 public class Solution {
     public int[] countBits(int num) {
         int[] result = new int[num + ];

         for (int i = ; i <= num; i++) {
             result[i] = countEach(i);
         }

         return result;
     }

     public int countEach(int num) {
         int result = ;

         while (num != ) {
             if ((num & ) == ) {
                 result++;
             }
             num = num >>> ;
         }

         return result;
     }
 }

2. Improved Solution

For number 2(10), 4(100), 8(1000), 16(10000), ..., the number of 1's is 1. Any other number can be converted to be 2^m + x. For example, 9=8+1, 10=8+2. The number of 1's for any other number is 1 + # of 1's in x.

     public int[] countBits(int num) {
         int[] result = new int[num + ];

         int pow = ;
         for (int i = ; i <= num; i++) {
             if (i == pow) {
                 result[i] = ;
                 pow = pow << ;
             } else {
                 int p = pow >> ;
                 result[i] = result[p] + result[i - p];
             }
         }
         return result;
     }