zoukankan      html  css  js  c++  java
  • LC 672. Bulb Switcher II

    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].

    Runtime: 4 ms, faster than 29.00% of C++ online submissions for Bulb Switcher II.

    找规律题。

    class Solution {
    public:
        int flipLights(int n, int m) {
          if(m == 0) return 1;
          if(n == 1) return 2;
          if(n == 2 && m == 1) return 3;
          if(n == 2) return 4;
          if(m==1) return 4;
          if(m == 2) return 7;
          return 8;
        }
    };
  • 相关阅读:
    CSS
    表单
    框架
    表格
    列表
    定位--position属性
    浮动
    选择结构
    数组
    TextView(标签控件)
  • 原文地址:https://www.cnblogs.com/ethanhong/p/10195577.html
Copyright © 2011-2022 走看看