338. Counting Bits Java Solutin

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 public class Solution {
 2     public int[] countBits(int num) {
 3         int[] result = new int[num+1];
 4         for(int i=0;i<num+1;i++){
 5             int count = 0;
 6             int j = i;
 7             while(j!=0){
 8                 j=j&(j-1);
 9                 count++;
10             }
11             result[i] = count;
12         }
13         return result;
14     }
15 }

public int[] countBits(int num) {for (int n = 0; n <= num; ++n) {
            // binaryNum接收转为二进制的数字
            String binaryNum = Integer.toBinaryString(n);
            char[] charArr = binaryNum.toCharArray();
            int i = 0;
            for (int m = 0; m < charArr.length; ++m) {
                if (charArr[m] == '1') {
                    i++;
                }
            }
            res[n] = i;
        }return res;
    }

第二个解法是网上看到，第一个解法根据编程之美书中求一个数中二进制表示中的 1 的位数得来。