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✨feat: add 672 #680

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Sep 15, 2022
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✨feat: add 672 #680

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2 changes: 1 addition & 1 deletion LeetCode/671-680/672. 灯泡开关 Ⅱ（中等）.md
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Expand Up @@ -68,7 +68,7 @@ Tag : 「脑筋急转弯」、「找规律」
* 当 $k = 0$ 时，无论 $n$ 为何值，都只有起始（全 `1`）一种状态；
* 当 $k > 0$ 时，根据 $n$ 进一步分情况讨论：
* 当 $n = 1$ 时，若 $k$ 为满足「$k > 0$」的最小值 $1$ 时，能够取满「`1`/`0`」两种情况，而其余更大 $k$ 值情况能够使用操作无效化（不影响灯的状态）；
* 当 $n = 2$ 时，若 $k = 1$，能够取得「`11`/`10`/`01`」三种状态，当 $k = 2$ 时，能够取满「`11`/`10`/`01`/`00`」四种状态，其余更大 $k$ 可以通过前 $k - 1$ 步归结到任一状态，再通过最后一次的操作 $1$ 归结到任意状态；
* 当 $n = 2$ 时，若 $k = 1$，能够取得「`00`/`10`/`01`」三种状态，当 $k = 2$ 时，能够取满「`11`/`10`/`01`/`00`」四种状态，其余更大 $k$ 可以通过前 $k - 1$ 步归结到任一状态，再通过最后一次的操作 $1$ 归结到任意状态；
* 当 $n = 3$ 时，若 $k = 1$ 时，对应 $4$ 种操作可取得 $4$ 种方案；当 $k = 2$ 时，可取得 $7$ 种状态；而当 $k = 3$ 时可取满 $2^3 = 8$ 种状态，更大的 $k$ 值可通过同样的方式归结到取满的 $8$ 种状态。
* 当 $n > 3$ 时，根据四类操作可知，灯泡每 $6$ 组一循环（对应序列 `k + 1`、`2k + 2`、`2k + 1` 和 `3k + 1`），即只需考虑 $n <= 6$ 的情况，而 $n = 4$、$n = 5$ 和 $n = 6$ 时，后引入的灯泡状态均不会产生新的组合（即新引入的灯泡状态由前三个灯泡的状态所唯一确定），因此均可归纳到 $n = 3$ 的情况。

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