Counting Bits

update Jul3, 2017 11:13

leetcode

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.

Basic Idea:

仔细观察二进制数的规律就会发现有如下关系：

`dp[i] = dp[i / 2] when i % 2 == 0`

`else dp[i] = dp[i / 2] + 1`

利用这点性质，我们就可以得出我们的 O(n) dp 算法。

Python Code：

    class Solution(object):
        def countBits(self, num):
            """
            :type num: int
            :rtype: List[int]
            """
            res = [0] * (num + 1)
            for i in range(1, len(res)):
                res[i] = res[i / 2] if i % 2 == 0 else res[i / 2] + 1
            return res