leetcode 338. counting bits

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.

####解题思路
一个数 * 2 就是把它的二进制形式全部左移一位，反过来看，一个数的二进制形式除去最后一位之外，其余位包含的1的总数，和其一半（/2）对应的二进制形式中的1的总数是相同的，也就是右移1位对应的结果。另外，最后1位是不是1，可以用i & 1来判断。

 public int[] countBits(int num) {
        int[] res = new int[num+1];
        for(int i=1;i<=num;i++){
            res[i] = res[i >> 1] + (i & 1);
        }
        return res;
    }
}