1006._Clumsy_Factorial.md

1006. Clumsy Factorial

难度: Medium

刷题内容

原题连接

• https://leetcode.com/problems/clumsy-factorial/

内容描述

Normally, the factorial of a positive integer n is the product of all positive integers less than or equal to n.  For example, factorial(10) = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1.

We instead make a clumsy factorial: using the integers in decreasing order, we swap out the multiply operations for a fixed rotation of operations: multiply (*), divide (/), add (+) and subtract (-) in this order.

For example, clumsy(10) = 10 * 9 / 8 + 7 - 6 * 5 / 4 + 3 - 2 * 1.  However, these operations are still applied using the usual order of operations of arithmetic: we do all multiplication and division steps before any addition or subtraction steps, and multiplication and division steps are processed left to right.

Additionally, the division that we use is floor division such that 10 * 9 / 8 equals 11.  This guarantees the result is an integer.

Implement the clumsy function as defined above: given an integer N, it returns the clumsy factorial of N.

 

Example 1:

Input: 4
Output: 7
Explanation: 7 = 4 * 3 / 2 + 1
Example 2:

Input: 10
Output: 12
Explanation: 12 = 10 * 9 / 8 + 7 - 6 * 5 / 4 + 3 - 2 * 1
 

Note:

1 <= N <= 10000
-2^31 <= answer <= 2^31 - 1  (The answer is guaranteed to fit within a 32-bit integer.)

解题方案

思路 1 - 时间复杂度: O(1)- 空间复杂度: O(1)******

自己在leetcode的第四篇post

We can easily observe below:

5 * 4 / 3 = 6
6 * 5 / 4 = 7
10 * 9 / 8 = 11
...
...
...

so we can get this formula: i * (i-1) / (i-1) = i+1 when i >= 5

so i * (i-1) / (i-2) + (i-3) = 2 * i - 2

we can simplify our computation as below:

    i * (i-1) / (i-2) + (i-3) - (i-4) * ....
=   2 * i - 2 - (i-4) * ....
=   2 * i - 2 - (2 * (i-4) - 2) ...
=   

we can call each 4 numbers a chunk, so from N % 4 we can know how many chunks there are, then the rest 0, 1, 3 and 4 elements will influence our final result.

After consider the corner case, we can arrive the solution:

class Solution:
    def clumsy(self, N: int) -> int:
        if N <= 2:
            return N
        if N <= 4:
            return N + 3
        
        if (N - 4) % 4 == 0:
            return N + 1
        elif (N - 4) % 4 <= 2:
            return N + 2
        else:
            return N - 1

class Solution:
    def clumsy(self, N: int) -> int:
        if N <= 2:
            return N
        if N <= 4:
            return N + 3
        
        if (N - 4) % 4 == 0:
            return N + 1
        elif (N - 4) % 4 <= 2:
            return N + 2
        else:
            return N - 1