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

  public class Solution {
      public int[] countBits(int num) {
          int[] result = new int[num+1];
          for(int i=0;i<num+1;i++){
              int count = 0;
              int j = i;
              while(j!=0){
                  j=j&(j-1);
                  count++;
              }
              result[i] = count;
          }
          return result;
      }
  }

  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。