Leet Code OJ 338. Counting Bits [Difficulty:easy]

Topic:
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 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 I do it in linear time O (n)/possibly in a single pass?
Space complexity should be O (n).
Can do it like a boss? Do it without the using any builtin function like __builtin_popcount in C + + or any other language.
Hint:
You should make the use of the produced already.

Translation:
Given a non-negative integer num, for each 0<=i<=num integer i, computes the number of the binary representation of I in 1, returning these numbers as an array.
For example, enter num = 5 and you should return [0,1,1,2,1,2].

Analysis:
It is easy to get the "Java Code 2" scheme in the usual way, but the time complexity of this scheme is O (NLOGN).
By analyzing the first 64 elements of an array (num=63), we find that the arrays are rendered in a certain regularity, repeating as shown:

Thus we find that 0112 is a basic element, which is repeated repeatedly and can be inferred: if the first element is known as result[0], then the second third element is result[0]+1, and the fourth element is result[0]+2, thus obtaining the first 4 elements result[0]~ RESULT[3]; Based on these 4 elements, we can get
Result[4]=result[0]+1,result[5]=result[1]+1 ...,
Result[8]=result[0]+1,result[9]=result[1]+1 ...,
Result[12]=result[0]+2,result[13]=result[1]+2 ...;
And so on, you get all the arrays.

Java version Code 1:

 Public  class solution {     Public int[]countbits(intNUM) {int[] result =New int[Num +1];intRange =1; result[0] =0;BooleanStop =false; while(!stop)            {stop = Fillnum (result, range); Range *=4; }returnResult } Public Boolean Fillnum(int[] Nums,intRange) { for(inti =0; I < range; i++) {if(Range + i < nums.length) {Nums[range + i] = Nums[i] +1; }Else{return true; }if(2* Range + I < nums.length) {nums[2* range + i] = Nums[i] +1; }if(3* Range + I < nums.length) {nums[3* range + i] = Nums[i] +2; }        }return false; }}

Java Version Code 2:

 Public classSolution { Public int[]countbits(intNUM) {int[] result=New int[num+1]; result[0]=0; for(intI=1; i<=num;i++) {result[i]=getcount (i); }returnResult } Public int GetCount(intNUM) {intCount=0; while(num!=0){if((num&1)==1) {count++; } num/=2; }returnCount }}

Leet Code OJ 338. Counting Bits [Difficulty:easy]