[Leetcode] 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.

public class Solution {    public int[] countBits(int num) {        int[] result = new int[num+1];        result[0] = 0;        for(int i = 1; i <= num; i++) {            // odd            if((i & 0x1) != 0) {                result[i] = result[i-1] + 1;            }            else {                int change = 1 - numOfRightmostBits(i-1);                result[i] = result[i-1] + change;            }        }        return result;    }        private int numOfRightmostBits(int num) {        int result = 0;        while((num & 0x1) != 0) {            result++;            num = num >> 1;        }        return result;    }}

public class Solution {    public int[] countBits(int num) {        int[] result = new int[num + 1];        int pow = 1;        int p = 1;        for(int i = 1; i <= num; i++) {            if(i == pow) {                result[i] = 1;                p = 1;                pow = pow << 1;            }            else {                result[i] = result[p] + 1;                p++;            }        }        return result;    }}