[LeetCode]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?

题解为：0~num中2进制中1的个数，用数组表示。

code:

public class Solution {
    public int[] countBits(int num) {
        
        int[] rets = new int[num+1];
        rets[0] = 0;
        int pow = 1;
        for(int i=1, t =0; i<=num; i++, t++)
        {
            if(pow == i)
            {
                pow *=2;
                t = 0;
            }
            rets[i] = rets[t] +1;
        }
        return rets;
    }
}