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.

Credits:
Special thanks to @ syedee for adding this problem and creating all test cases.

动态规划解题。动规的条件是f[i] = f[i / 2] + i % 2。把0-16的二进制形式写出来可以看出规律。代码如下：

public class Solution {
    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;
    }
}