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
= 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
= dp[n - offset] + 1
return dp