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 1:

Input: 2
Output: [0,1,1]

Example 2:

Input: 5
Output: [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.

这题妙在

十进制：             0 1 2 3      4 5 6 7    8 9 10 11 12 13 14 15    16... 

二进制中1的个数：0 1 1 2       1 2 2 3    1  2 2  3  2  3  3   4     1...

 

这题最妙在于当遇到2的pow时，把t置零，然后在前面数字二进制中1的个数的基础上加1

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