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

作者： 负雪明烛
id： fuxuemingzhu
个人博客： http://fuxuemingzhu.cn/

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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。

可以看http://blog.csdn.net/huanghanqian/article/details/77857912

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