【LeetCode】 338. 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.

Hint:

1. You should make use of what you have produced already.
2. Divide the numbers in ranges like [2-3], [4-7], [8-15] and so on. And try to generate new range from previous.
3. Or does the odd/even status of the number help you in calculating the number of 1s?

public class Solution {    public int[] countBits(int num) {        int[] res = new int[num + 1];        for (int i = 0; i <= num; i++) {            res[i] = Integer.bitCount(i);        }        return res;    }}

public class Solution {    public int[] countBits(int num) {        int result[] = new int[num + 1];        int offset = 1;        for (int index = 1; index < num + 1; ++index){            if (offset * 2 == index){                offset *= 2;            }            result[index] = result[index - offset] + 1;        }        return result;    }}