update: 添加问题“2147.分隔长廊的方案数”的代码和题解 (#1224)

LetMeFly666 · LetMeFly666 · commit 023153a4b1e7 · 2025-12-14T18:34:40.000+08:00
2147: AC.cpp (#1223) + word(en)早上 没有座位竟然是0种方案数，按理说一个隔板都不放不应该是唯一的一种方案吗。 cpp - AC,86.11%,100.00% 、AC,96.53%,100.00% Signed-off-by: LetMeFly666 <814114971@qq.com>
diff --git a/.commitmsg b/.commitmsg
@@ -1 +1,4 @@
-3606: AC.cpp - AC,59.14%,92.47% (#1221) + word(en)
+2147: AC.cpp (#1223) + word(en)早上
+
+没有座位竟然是0种方案数，按理说一个隔板都不放不应该是唯一的一种方案吗。
+cpp - AC,86.11%,100.00% 、AC,96.53%,100.00%
diff --git a/Codes/2147-number-of-ways-to-divide-a-long-corridor.cpp b/Codes/2147-number-of-ways-to-divide-a-long-corridor.cpp
@@ -0,0 +1,45 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-12-14 17:30:12
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-12-14 17:37:56
+ */
+#if defined(_WIN32) || defined(__APPLE__)
+#include "_[1,2]toVector.h"
+#endif
+
+typedef long long ll;
+const ll MOD = 1e9 + 7;
+class Solution {
+public:
+    int numberOfWays(string& corridor) {
+        ll ans = 1;
+        int now = 0;
+        bool ing = false;  // 正在处理两块座位之间的绿植
+        bool atLeast2 = false;
+        for (char c : corridor) {
+            if (c == 'S') {
+                if (ing) {
+                    ans = ans * (now + 1) % MOD;
+                    ing = false;
+                    now = 1;
+                } else {
+                    now++;
+                    if (now == 2) {
+                        ing = true;
+                        now = 0;
+                        atLeast2 = true;
+                    }
+                }
+            } else {  // 'P'
+                if (ing) {
+                    now++;
+                }
+            }
+        }
+        if (!ing && now != 0 || !atLeast2) {
+            return 0;
+        }
+        return static_cast<int>(ans);
+    }
+};
diff --git a/Codes/2147-number-of-ways-to-divide-a-long-corridor_Better.cpp b/Codes/2147-number-of-ways-to-divide-a-long-corridor_Better.cpp
@@ -0,0 +1,45 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-12-14 18:28:41
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-12-14 18:29:02
+ */
+#if defined(_WIN32) || defined(__APPLE__)
+#include "_[1,2]toVector.h"
+#endif
+
+typedef long long ll;
+const ll MOD = 1e9 + 7;
+class Solution {
+public:
+    int numberOfWays(string& corridor) {
+        ll ans = 1;
+        int now = 0;
+        bool ing = false;  // 正在处理两块座位之间的绿植
+        bool atLeast2 = false;
+        for (char c : corridor) {
+            if (c == 'S') {
+                if (ing) {
+                    ans = ans * (now + 1) % MOD;
+                    ing = false;
+                    now = 1;
+                } else {
+                    now++;
+                    if (now == 2) {
+                        ing = true;
+                        now = 0;
+                        atLeast2 = true;
+                    }
+                }
+            } else {  // 'P'
+                if (ing) {
+                    now++;
+                }
+            }
+        }
+        if (!ing || !atLeast2) {
+            return 0;
+        }
+        return static_cast<int>(ans);
+    }
+};
diff --git a/Codes/3606-coupon-code-validator.cpp b/Codes/3606-coupon-code-validator.cpp
@@ -1,14 +1,13 @@
 /*
  * @Author: LetMeFly
- * @Date: 2025-12-13 22:26:53
+ * @Date: 2025-12-13 22:41:43
  * @LastEditors: LetMeFly.xyz
- * @LastEditTime: 2025-12-13 22:42:38
+ * @LastEditTime: 2025-12-13 22:42:29
  */
 #if defined(_WIN32) || defined(__APPLE__)
 #include "_[1,2]toVector.h"
 #endif
 
-// THIS CANNOT PASS
 class Solution {
 private:
     inline bool is_ok(string& s) {
@@ -29,13 +28,13 @@ class Solution {
             if (!is_ok(code[i])) {
                 continue;
             }
-            if (businessLine[i][0] == 'e') {
+            if (businessLine[i] == "electronics") {
                 electronics.push_back(code[i]);
-            } else if (businessLine[i][0] == 'g') {
+            } else if (businessLine[i] == "grocery") {
                 grocery.push_back(code[i]);
-            } else if (businessLine[i][0] == 'p') {
+            } else if (businessLine[i] == "pharmacy") {
                 pharmacy.push_back(code[i]);
-            } else {
+            } else if (businessLine[i] == "restaurant") {
                 restaurant.push_back(code[i]);
             }
         }
diff --git a/Codes/3606-coupon-code-validator_AC.cpp b/Codes/3606-coupon-code-validator_AC.cpp
diff --git a/README.md b/README.md
@@ -712,6 +712,7 @@
 |2132.用邮票贴满网格图|困难|<a href="https://leetcode.cn/problems/stamping-the-grid/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2023/12/14/LeetCode%202132.%E7%94%A8%E9%82%AE%E7%A5%A8%E8%B4%B4%E6%BB%A1%E7%BD%91%E6%A0%BC%E5%9B%BE/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/135002925" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/stamping-the-grid/solutions/2566644/letmefly-2132yong-you-piao-tie-man-wang-znigh/" target="_blank">LeetCode题解</a>|
 |2138.将字符串拆分为若干长度为k的组|简单|<a href="https://leetcode.cn/problems/divide-a-string-into-groups-of-size-k/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2025/06/22/LeetCode%202138.%E5%B0%86%E5%AD%97%E7%AC%A6%E4%B8%B2%E6%8B%86%E5%88%86%E4%B8%BA%E8%8B%A5%E5%B9%B2%E9%95%BF%E5%BA%A6%E4%B8%BAk%E7%9A%84%E7%BB%84/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/148844890" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/divide-a-string-into-groups-of-size-k/solutions/3706437/letmefly-2138jiang-zi-fu-chuan-chai-fen-acsxz/" target="_blank">LeetCode题解</a>|
 |2140.解决智力问题|中等|<a href="https://leetcode.cn/problems/solving-questions-with-brainpower/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2025/04/03/LeetCode%202140.%E8%A7%A3%E5%86%B3%E6%99%BA%E5%8A%9B%E9%97%AE%E9%A2%98/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/146991317" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/solving-questions-with-brainpower/solutions/3639601/letmefly-2140jie-jue-zhi-li-wen-ti-ji-yi-pxc4/" target="_blank">LeetCode题解</a>|
+|2147.分隔长廊的方案数|困难|<a href="https://leetcode.cn/problems/number-of-ways-to-divide-a-long-corridor/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2025/12/14/LeetCode%202147.%E5%88%86%E9%9A%94%E9%95%BF%E5%BB%8A%E7%9A%84%E6%96%B9%E6%A1%88%E6%95%B0/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/155916226" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/number-of-ways-to-divide-a-long-corridor/solutions/3856839/letmefly-2147fen-ge-chang-lang-de-fang-a-tn48/" target="_blank">LeetCode题解</a>|
 |2171.拿出最少数目的魔法豆|中等|<a href="https://leetcode.cn/problems/removing-minimum-number-of-magic-beans/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2024/01/18/LeetCode%202171.%E6%8B%BF%E5%87%BA%E6%9C%80%E5%B0%91%E6%95%B0%E7%9B%AE%E7%9A%84%E9%AD%94%E6%B3%95%E8%B1%86/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/135682601" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/removing-minimum-number-of-magic-beans/solutions/2609831/letmefly-2171na-chu-zui-shao-shu-mu-de-m-jima/" target="_blank">LeetCode题解</a>|
 |2176.统计数组中相等且可以被整除的数对|简单|<a href="https://leetcode.cn/problems/count-equal-and-divisible-pairs-in-an-array/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2025/04/17/LeetCode%202176.%E7%BB%9F%E8%AE%A1%E6%95%B0%E7%BB%84%E4%B8%AD%E7%9B%B8%E7%AD%89%E4%B8%94%E5%8F%AF%E4%BB%A5%E8%A2%AB%E6%95%B4%E9%99%A4%E7%9A%84%E6%95%B0%E5%AF%B9/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/147315213" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/count-equal-and-divisible-pairs-in-an-array/solutions/3654492/letmefly-2176tong-ji-shu-zu-zhong-xiang-643mr/" target="_blank">LeetCode题解</a>|
 |2178.拆分成最多数目的正偶数之和|中等|<a href="https://leetcode.cn/problems/maximum-split-of-positive-even-integers/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2023/07/06/LeetCode%202178.%E6%8B%86%E5%88%86%E6%88%90%E6%9C%80%E5%A4%9A%E6%95%B0%E7%9B%AE%E7%9A%84%E6%AD%A3%E5%81%B6%E6%95%B0%E4%B9%8B%E5%92%8C/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/131567568" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/maximum-split-of-positive-even-integers/solutions/2332513/letmefly-2178chai-fen-cheng-zui-duo-shu-m44er/" target="_blank">LeetCode题解</a>|
diff --git a/Solutions/LeetCode 2147.分隔长廊的方案数.md b/Solutions/LeetCode 2147.分隔长廊的方案数.md
@@ -0,0 +1,136 @@
+---
+title: 2147.分隔长廊的方案数：非Hard组合数学
+date: 2025-12-14 18:20:28
+tags: [题解, LeetCode, 困难, 数学, 组合数学, 排列组合, 字符串, 动态规划]
+categories: [题解, LeetCode]
+index_img: https://assets.leetcode.com/uploads/2021/12/04/1.png
+---
+
+# 【LetMeFly】2147.分隔长廊的方案数：非Hard组合数学
+
+力扣题目链接：[https://leetcode.cn/problems/number-of-ways-to-divide-a-long-corridor/](https://leetcode.cn/problems/number-of-ways-to-divide-a-long-corridor/)
+
+<p>在一个图书馆的长廊里，有一些座位和装饰植物排成一列。给你一个下标从 <strong>0</strong>&nbsp;开始，长度为 <code>n</code>&nbsp;的字符串&nbsp;<code>corridor</code>&nbsp;，它包含字母&nbsp;<code>'S'</code> 和&nbsp;<code>'P'</code>&nbsp;，其中每个&nbsp;<code>'S'</code>&nbsp;表示一个座位，每个&nbsp;<code>'P'</code>&nbsp;表示一株植物。</p>
+
+<p>在下标 <code>0</code>&nbsp;的左边和下标 <code>n - 1</code>&nbsp;的右边 <strong>已经</strong>&nbsp;分别各放了一个屏风。你还需要额外放置一些屏风。每一个位置&nbsp;<code>i - 1</code> 和&nbsp;<code>i</code>&nbsp;之间（<code>1 &lt;= i &lt;= n - 1</code>），至多能放一个屏风。</p>
+
+<p>请你将走廊用屏风划分为若干段，且每一段内都 <strong>恰好有两个座位</strong>&nbsp;，而每一段内植物的数目没有要求。可能有多种划分方案，如果两个方案中有任何一个屏风的位置不同，那么它们被视为 <strong>不同</strong> 方案。</p>
+
+<p>请你返回划分走廊的方案数。由于答案可能很大，请你返回它对&nbsp;<code>10<sup>9</sup> + 7</code>&nbsp;<strong>取余</strong>&nbsp;的结果。如果没有任何方案，请返回&nbsp;<code>0</code>&nbsp;。</p>
+
+<p>&nbsp;</p>
+
+<p><strong>示例 1：</strong></p>
+
+<p><img alt="" src="https://assets.leetcode.com/uploads/2021/12/04/1.png" style="width: 410px; height: 199px;"></p>
+
+<pre><b>输入：</b>corridor = "SSPPSPS"
+<b>输出：</b>3
+<b>解释：</b>总共有 3 种不同分隔走廊的方案。
+上图中黑色的竖线表示已经放置好的屏风。
+上图每种方案中，每一段都恰好有 <strong>两个</strong>&nbsp;座位。
+</pre>
+
+<p><strong>示例 2：</strong></p>
+
+<p><img alt="" src="https://assets.leetcode.com/uploads/2021/12/04/2.png" style="width: 357px; height: 68px;"></p>
+
+<pre><b>输入：</b>corridor = "PPSPSP"
+<b>输出：</b>1
+<b>解释：</b>只有 1 种分隔走廊的方案，就是不放置任何屏风。
+放置任何的屏风都会导致有一段无法恰好有 2 个座位。
+</pre>
+
+<p><strong>示例 3：</strong></p>
+
+<p><img alt="" src="https://assets.leetcode.com/uploads/2021/12/12/3.png" style="width: 115px; height: 68px;"></p>
+
+<pre><b>输入：</b>corridor = "S"
+<b>输出：</b>0
+<b>解释：</b>没有任何方案，因为总是有一段无法恰好有 2 个座位。
+</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>n == corridor.length</code></li>
+	<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
+	<li><code>corridor[i]</code>&nbsp;要么是&nbsp;<code>'S'</code>&nbsp;，要么是&nbsp;<code>'P'</code> 。</li>
+</ul>
+
+
+    
+## 解题方法：遍历
+
+从左往右遍历，每出现总计两个座位就要进行一次分隔，本次分隔方案数为两个座位中后一个座位与**下一个**座位之间的绿植数加一。（若后续再无座位，则不需要放置隔板）
+
+总方案数就是每次放置隔板时的方案数之积。
+
+额外注意，本题给定的答案中，若没有座位，则输出0而非“一个隔板都不放的这唯一一种方案”1。
+
+### 具体做法
+
+使用一个变量ing记录当前是否处在两个座位之后下一个座位之前的数绿植状态，使用一个变量now记录当前总计座位数或绿植数，遍历时候使用几个if-else就好了。
+
+额外注意，可以使用一个变量atLeast2记录是否至少有两个座位，遍历过程中一旦出现累计两个座位则将该值赋值为true。
+
+最终若不是在数绿植状态（不是刚好两个座位后）或一共也没有两个座位，返回0。
+
+### 时空复杂度分析
+
++ 时间复杂度$O(len(corridor))$
++ 空间复杂度$O(1)$
+
+### AC代码
+
+#### C++
+
+```cpp
+/*
+ * @LastEditTime: 2025-12-14 17:37:56
+ */
+typedef long long ll;
+const ll MOD = 1e9 + 7;
+class Solution {
+public:
+    int numberOfWays(string& corridor) {
+        ll ans = 1;
+        int now = 0;
+        bool ing = false;  // 正在处理两块座位之间的绿植
+        bool atLeast2 = false;
+        for (char c : corridor) {
+            if (c == 'S') {
+                if (ing) {
+                    ans = ans * (now + 1) % MOD;
+                    ing = false;
+                    now = 1;
+                } else {
+                    now++;
+                    if (now == 2) {
+                        ing = true;
+                        now = 0;
+                        atLeast2 = true;
+                    }
+                }
+            } else {  // 'P'
+                if (ing) {
+                    now++;
+                }
+            }
+        }
+        if (!ing || !atLeast2) {
+            return 0;
+        }
+        return static_cast<int>(ans);
+    }
+};
+```
+
+* 执行用时分布8ms击败96.53%
+* 消耗内存分布26.95MB击败100.00%
+
+> 同步发文于[CSDN](https://letmefly.blog.csdn.net/article/details/155916226)和我的[个人博客](https://blog.letmefly.xyz/)，原创不易，转载经作者同意后请附上[原文链接](https://blog.letmefly.xyz/2025/12/14/LeetCode%202147.%E5%88%86%E9%9A%94%E9%95%BF%E5%BB%8A%E7%9A%84%E6%96%B9%E6%A1%88%E6%95%B0/)哦~
+>
+> 千篇源码题解[已开源](https://github.com/LetMeFly666/LeetCode)
diff --git a/Solutions/LeetCode 3531.统计被覆盖的建筑.md b/Solutions/LeetCode 3531.统计被覆盖的建筑.md
@@ -3,6 +3,7 @@ title: 3531.统计被覆盖的建筑：最大最小值
 date: 2025-12-11 18:39:45
 tags: [题解, LeetCode, 中等, 数组, 排序]
 categories: [题解, LeetCode]
+index_img: https://pic.leetcode.cn/1745660407-qtNUjI-telegram-cloud-photo-size-5-6212982906394101085-m.jpg
 ---
 
 # 【LetMeFly】3531.统计被覆盖的建筑：最大最小值
diff --git a/Solutions/Other-English-LearningNotes-SomeWords.md b/Solutions/Other-English-LearningNotes-SomeWords.md
@@ -1666,6 +1666,11 @@ categories: [自用]
 |dome|n. 穹顶，圆屋顶，穹顶建筑物|
 |||
 |mercantile|adj. 商业化，贸易的|
+|||
+|flannel|n. 法兰绒|
+|seaman|n. 海员，水兵，水手|
+
+
 
 + 这个web要是能设计得可以闭眼(完全不睁眼)键盘控制背单词就好了。
 + 也许可以加个AI用最近词编故事功能(返回接口中支持标注所使用单词高亮?)
