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【LeetCode】672. Bulb Switcher II 解题报告(Python)

凌得涂 2022-02-10 阅读 40


id: fuxuemingzhu

个人博客: ​​​http://fuxuemingzhu.cn/​​​


目录


  • ​​题目描述​​
  • ​​题目大意​​
  • ​​解题方法​​
  • ​​日期​​


题目地址:​​https://leetcode.com/problems/bulb-switcher-ii/description/​​

题目描述

There is a room with n lights which are turned on initially and 4 buttons on the wall. After performing exactly m unknown operations towards buttons, you need to return how many different kinds of status of the n lights could be.

Suppose n lights are labeled as number [1, 2, 3 …, n], function of these 4 buttons are given below:


  1. Flip all the lights.
  2. Flip lights with even numbers.
  3. Flip lights with odd numbers.
  4. Flip lights with (3k + 1) numbers, k = 0, 1, 2, …

Example 1:

Input: n = 1, m = 1.
Output: 2
Explanation: Status can be: [on], [off]

Example 2:

Input: n = 2, m = 1.
Output: 3
Explanation: Status can be: [on, off], [off, on], [off, off]

Example 3:

Input: n = 3, m = 1.
Output: 4
Explanation: Status can be: [off, on, off], [on, off, on], [off, off, off], [off, on, on].

Note: n and m both fit in range [0, 1000].

题目大意

有四种操作。分别对灯进行变换。。不是一个数学题,这个是脑筋急转弯。

解题方法

额,不会做。脑子不够用。。这个题被踩了这么多下,其实没有做的必要了。


代码:

class Solution:
def flipLights(self, n, m):
"""
:type n: int
:type m: int
:rtype: int
"""
if m == 0:
return 1
if n >= 3:
return 4 if m == 1 else 7 if m == 2 else 8
if n == 2:
return 3 if m == 1 else 4
if n == 1:
return 2

日期

2018 年 3 月 5 日



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