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 + 1];

        for (int i = 0; i <= num; i++) {
            result[i] = countEach(i);
        }

        return result;
    }

    public int countEach(int num) {
        int result = 0;

        while (num != 0) {
            if ((num & 1) == 1) {
                result++;
            }
            num = num >>> 1;
        }

        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 + 1];

        int pow = 1;
        for (int i = 1; i <= num; i++) {
            if (i == pow) {
                result[i] = 1;
                pow = pow << 1;
            } else {
                int p = pow >> 1;
                result[i] = result[p] + result[i - p];
            }
        }
        return result;
    }