338. Counting Bits -Medium

Question

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.

给出一个非负整数num，对[0, num]范围内的数分别计算它们的二进制数中1的个数，用数组形式返回。

Example

For num = 5 you should return [0,1,1,2,1,2].

Solution

• 动态规划解。其实这是道找规律的题目，我们首先列出0-15的数对应的二进制中1的个数

  数字 二进制中1的个数 递推关系式 0 0 dp[0] = 0 1 1 dp[1] = dp[1-1] + 1 2 1 dp[2] = dp[2-2] + 1 3 2 dp[3] = dp[3-2] + 1 4 1 dp[4] = dp[4-4] + 1 5 2 dp[5] = dp[5-4] + 1 6 2 dp[6] = dp[6-4] + 1 7 3 dp[7] = dp[7-4] + 1 8 1 dp[8] = dp[8-8] + 1 9 2 dp[9] = dp[9-8] + 1 10 2 dp[10] = dp[10-8] + 1 11 3 dp[11] = dp[11-8] + 1 12 2 dp[12] = dp[12-8] + 1 13 3 dp[13] = dp[13-8] + 1 14 3 dp[14] = dp[14-8] + 1 15 4 dp[15] = dp[15-8] + 1

  综上，递推关系式为 dp[n] = dp[n - offset] + 1 (这个规律还真不怎么好找)，而offset的更新规律为，每当 offset * 2等于n时，offset就需要更新，即乘以2.

  class Solution(object):    def countBits(self, num):        """        :type num: int        :rtype: List[int]        """        # 0-num有num + 1个数        dp = [0] * (num + 1)        offset = 1        for n in range(1, num + 1):            # 根据规律，只要index = 2 * offset，offset需要乘以2            if offset * 2 == n:                offset *= 2            # dp[index] = dp[index - offset] + 1            dp[n] = dp[n - offset] + 1        return dp