zoukankan      html  css  js  c++  java
  • [Swift]LeetCode672. 灯泡开关 Ⅱ | Bulb Switcher II

    ★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
    ➤微信公众号:山青咏芝(shanqingyongzhi)
    ➤博客园地址:山青咏芝(https://www.cnblogs.com/strengthen/
    ➤GitHub地址:https://github.com/strengthen/LeetCode
    ➤原文地址:https://www.cnblogs.com/strengthen/p/10496919.html 
    ➤如果链接不是山青咏芝的博客园地址,则可能是爬取作者的文章。
    ➤原文已修改更新!强烈建议点击原文地址阅读!支持作者!支持原创!
    ★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

    There is a room with n lights which are turned on initially and 4 buttons on the wall. After performing exactly munknown 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].


    现有一个房间,墙上挂有 n 只已经打开的灯泡和 4 个按钮。在进行了 m 次未知操作后,你需要返回这 n 只灯泡可能有多少种不同的状态。

    假设这 n 只灯泡被编号为 [1, 2, 3 ..., n],这 4 个按钮的功能如下:

    1. 将所有灯泡的状态反转(即开变为关,关变为开)
    2. 将编号为偶数的灯泡的状态反转
    3. 将编号为奇数的灯泡的状态反转
    4. 将编号为 3k+1 的灯泡的状态反转(k = 0, 1, 2, ...)

    示例 1:

    输入: n = 1, m = 1.
    输出: 2
    说明: 状态为: [开], [关]
    

    示例 2:

    输入: n = 2, m = 1.
    输出: 3
    说明: 状态为: [开, 关], [关, 开], [关, 关]
    

    示例 3:

    输入: n = 3, m = 1.
    输出: 4
    说明: 状态为: [关, 开, 关], [开, 关, 开], [关, 关, 关], [关, 开, 开].
    

    注意: n 和 m 都属于 [0, 1000].


    Runtime: 4 ms
    Memory Usage: 18.9 MB
    1 class Solution {
    2     func flipLights(_ n: Int, _ m: Int) -> Int {
    3         var n = min(n, 3)
    4         return min(1 << n, 1 + m * n)
    5     }
    6 }

    4ms

     1 class Solution {
     2     struct State: Hashable, OptionSet {
     3         let rawValue: Int
     4 
     5         static let oddSensitive = State(rawValue: 1)
     6         static let evenInsensitive = State(rawValue: 2)
     7         static let oddInsensitive = State(rawValue: 4)
     8         static let evenSensitive = State(rawValue: 8)
     9         
    10         static let odd: State = [.oddSensitive, .oddInsensitive]
    11         static let even: State = [.evenSensitive, .evenInsensitive]
    12         static let sensitive: State = [.evenSensitive, .oddSensitive]
    13         
    14         static let all: State = [.oddSensitive, .oddInsensitive, .evenSensitive, .evenInsensitive]
    15     }
    16     
    17     func flipLights(_ n: Int, _ m: Int) -> Int {
    18         let mask = State(rawValue: 1 << min(n, 4) - 1)
    19         let permittedCount = Set<Int>(0...4).filter { $0 <= m && ($0 % 2) == (m % 2) }
    20 
    21         var states = Set<State>()
    22         
    23         for i in 0..<16 {
    24             var count = 0, state = State.all
    25             if (i & 1) != 0 {
    26                 state.formSymmetricDifference(.all)
    27                 count += 1
    28             }
    29             if (i & 2) != 0 {
    30                 state.formSymmetricDifference(.even)
    31                 count += 1
    32             }
    33             if (i & 4) != 0 {
    34                 state.formSymmetricDifference(.odd)
    35                 count += 1
    36             }
    37             if (i & 8) != 0 {
    38                 state.formSymmetricDifference(.sensitive)
    39                 count += 1
    40             }
    41             
    42             if permittedCount.contains(count) {
    43                 states.insert(state.intersection(mask))
    44             }
    45         }
    46         
    47         return states.count
    48     }
    49 }
  • 相关阅读:
    【特效】导航下拉菜单(二级三级都有)
    【特效】移入显示移出隐藏
    【特效】jquery选项卡插件,页面多个选项卡统一调用
    使用MR求解多个矩阵的乘积之后
    由SequenceFile.Writer(key,value)谈toString()方法
    自定义数据类型写入SequenceFile并读出
    hadoop中URI理解
    输入格式CombineFileInput
    分布式缓存DistributedCache的使用
    输入格式MultipleInput
  • 原文地址:https://www.cnblogs.com/strengthen/p/10496919.html
Copyright © 2011-2022 走看看