672. Bulb Switcher II

时间:2020-03-12
本文章向大家介绍672. Bulb Switcher II,主要包括672. Bulb Switcher II使用实例、应用技巧、基本知识点总结和需要注意事项,具有一定的参考价值,需要的朋友可以参考一下。

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].

class Solution {
    public int flipLights(int n, int m) {
        if (m == 0) return 1;
        if (n <= 0 || m < 0) return 0;
        
        if (n == 1) return 2;
        else if (n == 2) return (m == 1) ? 3 : 4;
        else return (m == 1) ? 4 : ((m == 2) ? 7 : 8);
    }
}

https://leetcode.com/problems/bulb-switcher-ii/discuss/107272/Short-and-Clean-Java-O(1)-Solution

。。无语

原文地址:https://www.cnblogs.com/wentiliangkaihua/p/12468745.html