Leetcode 338. Counting Bits

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 you 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.

Ideas:

1. Check what each bit is, which is what it says very simple idea

2. Even words, is 1 of the previous number of digits + 1, even words/2 until it is odd, because the two cups relative to the shift, 1 of the same number of digits

Code: C + + idea 2:

classSolution { Public: Vector<int> Countbits (intnum) {Vector<int>solve; if(Num <0)            returnsolve; Solve.push_back (0); if(num = =0){        returnsolve; }                 for(inti =1; I <= num;i++){            intindex =i; ifI2){//i is an odd numberSolve.push_back (Solve.back () +1); }            Else{                 while(index%2==0) Index/=2;            Solve.push_back (Solve[index]); }        }        returnsolve; }};

Leetcode 338. Counting Bits