微信开发中 ACCESS TOKEN 过期失效的解决方案详解
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2022-11-08
[leetcode] 672. Bulb Switcher II
Description
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:
Flip all the lights.Flip lights with even numbers.Flip lights with odd numbers.Flip lights with (3k + 1) numbers, k = 0, 1, 2, …Example 1:
Input: n = 1, m = 1.Output: 2Explanation: Status can be: [on], [off]
Example 2:
Input: n = 2, m = 1.Output: 3Explanation: Status can be: [on, off], [off, on], [off, off]
Example 3:
Input: n = 3, m = 1.Output: 4Explanation: 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].
分析
题目的意思是: 有n个灯,我按照规则做m次开关操作,求这n个灯最后的种类数
这道题最多只有8种情况(我不知道怎么的出来的),所以很适合分情况来讨论:
当m和n其中有任意一个数是0时,返回1当n = 1时,只有两种情况,0和1当n = 2时,这时候要看m的次数,如果m = 1,那么有三种状态 00,01,10当m > 1时,那么有四种状态,00,01,10,11当m = 1时,此时n至少为3,那么我们有四种状态,000,010,101,011当m = 2时,此时n至少为3,我们有七种状态:111,101,010,100,000,001,110当m > 2时,此时n至少为3,我们有八种状态:111,101,010,100,000,001,110,011
class Solution {public: int flipLights(int n, int m) { if(m==0||n==0) return 1; if(n==1) return 2; if(n==2) return m==1 ? 3:4; if(m==1) return 4; return m==2 ? 7:8; }};
参考文献
[LeetCode] Bulb Switcher II 灯泡开关之二
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