feat: update lc problems

Author: yanglbme

solution/0100-0199/0134.Gas Station/README.md

@@ -57,10 +57,10 @@ tags:
 <p><strong>提示:</strong></p>
 
 <ul>
-	<li><code>gas.length == n</code></li>
-	<li><code>cost.length == n</code></li>
+	<li><code>n == gas.length == cost.length</code></li>
 	<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
 	<li><code>0 &lt;= gas[i], cost[i] &lt;= 10<sup>4</sup></code></li>
+	<li>输入保证答案唯一。</li>
 </ul>
 
 <!-- description:end -->

solution/0100-0199/0134.Gas Station/README_EN.md

@@ -60,6 +60,7 @@ Therefore, you can't travel around the circuit once no matter where you star
 	<li><code>n == gas.length == cost.length</code></li>
 	<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
 	<li><code>0 &lt;= gas[i], cost[i] &lt;= 10<sup>4</sup></code></li>
+	<li>The input is generated such that the answer is unique.</li>
 </ul>
 
 <!-- description:end -->

solution/0100-0199/0195.Tenth Line/README_EN.md

@@ -23,41 +23,26 @@ tags:
 <p>Assume that <code>file.txt</code> has the following content:</p>
 
 <pre>
-
 Line 1
-
 Line 2
-
 Line 3
-
 Line 4
-
 Line 5
-
 Line 6
-
 Line 7
-
 Line 8
-
 Line 9
-
 Line 10
-
 </pre>
 
 <p>Your script should output the tenth line, which is:</p>
 
 <pre>
-
 Line 10
-
 </pre>
 
 <div class="spoilers"><b>Note:</b><br />
-
 1. If the file contains less than 10 lines, what should you output?<br />
-
 2. There&#39;s at least three different solutions. Try to explore all possibilities.</div>
 
 <!-- description:end -->

solution/0200-0299/0262.Trips and Users/README.md

@@ -36,8 +36,6 @@ id 是这张表的主键（具有唯一值的列）。
 status 是一个表示行程状态的枚举类型，枚举成员为(‘completed’, ‘cancelled_by_driver’, ‘cancelled_by_client’) 。
 </pre>
 
-<p>&nbsp;</p>
-
 <div class="original__bRMd">
 <div>
 <p>表：<code>Users</code></p>
@@ -57,8 +55,6 @@ users_id 是这张表的主键（具有唯一值的列）。
 banned 是一个表示用户是否被禁止的枚举类型，枚举成员为 (‘Yes’, ‘No’) 。
 </pre>
 
-<p>&nbsp;</p>
-
 <p><strong>取消率</strong> 的计算方式如下：(被司机或乘客取消的非禁止用户生成的订单数量) / (非禁止用户生成的订单总数)。</p>
 
 <p>编写解决方案找出&nbsp;<code>"2013-10-01"</code><strong>&nbsp;</strong>至&nbsp;<code>"2013-10-03"</code><strong>&nbsp;</strong>期间有&nbsp;<strong>至少&nbsp;</strong>一次行程的非禁止用户（<strong>乘客和司机都必须未被禁止</strong>）的 <strong>取消率</strong>。非禁止用户即 banned 为 No 的用户，禁止用户即 banned 为 Yes 的用户。其中取消率 <code>Cancellation Rate</code> 需要四舍五入保留 <strong>两位小数</strong> 。</p>

solution/0200-0299/0262.Trips and Users/README_EN.md

@@ -34,7 +34,7 @@ The table holds all taxi trips. Each trip has a unique id, while client_id and d
 Status is an ENUM (category) type of ('completed', 'cancelled_by_driver', 'cancelled_by_client').
 </pre>
 
-<p>&nbsp;</p>
+<p> </p>
 
 <p>Table: <code>Users</code></p>
 
@@ -51,15 +51,15 @@ The table holds all users. Each user has a unique users_id, and role is an ENUM
 banned is an ENUM (category) type of (&#39;Yes&#39;, &#39;No&#39;).
 </pre>
 
-<p>&nbsp;</p>
+<p> </p>
 
 <p>The <strong>cancellation rate</strong> is computed by dividing the number of canceled (by client or driver) requests with unbanned users by the total number of requests with unbanned users on that day.</p>
 
 <p>Write a solution to find the <strong>cancellation rate</strong> of requests with unbanned users (<strong>both client and driver must not be banned</strong>) each day between <code>&quot;2013-10-01&quot;</code> and <code>&quot;2013-10-03&quot;</code> with <strong>at least</strong> one trip. Round <code>Cancellation Rate</code> to <strong>two decimal</strong> points.</p>
 
 <p>Return the result table in <strong>any order</strong>.</p>
 
-<p>The&nbsp;result format is in the following example.</p>
+<p>The result format is in the following example.</p>
 
 <p>&nbsp;</p>
 <p><strong class="example">Example 1:</strong></p>

solution/0500-0599/0596.Classes More Than 5 Students/README.md

@@ -0,0 +1,92 @@
+---
+comments: true
+difficulty: 简单
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/0500-0599/0596.Classes%20More%20Than%205%20Students/README.md
+tags:
+    - 数据库
+---
+
+<!-- problem:start -->
+
+# [596. 超过 5 名学生的课](https://leetcode.cn/problems/classes-more-than-5-students)
+
+[English Version](/solution/0500-0599/0596.Classes%20More%20Than%205%20Students/README_EN.md)
+
+## 题目描述
+
+<!-- description:start -->
+
+<p>表:&nbsp;<code>Courses</code></p>
+
+<pre>
++-------------+---------+
+| Column Name | Type    |
++-------------+---------+
+| student     | varchar |
+| class       | varchar |
++-------------+---------+
+(student, class)是该表的主键（不同值的列的组合）。
+该表的每一行表示学生的名字和他们注册的班级。
+</pre>
+
+<p>&nbsp;</p>
+
+<p>查询&nbsp;<strong>至少有 5 个学生</strong> 的所有班级。</p>
+
+<p>以 <strong>任意顺序 </strong>返回结果表。</p>
+
+<p>结果格式如下所示。</p>
+
+<p>&nbsp;</p>
+
+<p><strong class="example">示例 1:</strong></p>
+
+<pre>
+<strong>输入:</strong> 
+Courses table:
++---------+----------+
+| student | class    |
++---------+----------+
+| A       | Math     |
+| B       | English  |
+| C       | Math     |
+| D       | Biology  |
+| E       | Math     |
+| F       | Computer |
+| G       | Math     |
+| H       | Math     |
+| I       | Math     |
++---------+----------+
+<strong>输出:</strong> 
++---------+ 
+| class &nbsp; | 
++---------+ 
+| Math &nbsp; &nbsp;| 
++---------+
+<strong>解释: </strong>
+-数学课有 6 个学生，所以我们包括它。
+-英语课有 1 名学生，所以我们不包括它。
+-生物课有 1 名学生，所以我们不包括它。
+-计算机课有 1 个学生，所以我们不包括它。</pre>
+
+<!-- description:end -->
+
+## 解法
+
+<!-- solution:start -->
+
+### 方法一
+
+<!-- tabs:start -->
+
+#### MySQL
+
+```sql
+
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- problem:end -->

solution/0500-0599/0596.Classes More Than 5 Students/README_EN.md

@@ -0,0 +1,92 @@
+---
+comments: true
+difficulty: Easy
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/0500-0599/0596.Classes%20More%20Than%205%20Students/README_EN.md
+tags:
+    - Database
+---
+
+<!-- problem:start -->
+
+# [596. Classes More Than 5 Students](https://leetcode.com/problems/classes-more-than-5-students)
+
+[中文文档](/solution/0500-0599/0596.Classes%20More%20Than%205%20Students/README.md)
+
+## Description
+
+<!-- description:start -->
+
+<p>Table: <code>Courses</code></p>
+
+<pre>
++-------------+---------+
+| Column Name | Type    |
++-------------+---------+
+| student     | varchar |
+| class       | varchar |
++-------------+---------+
+(student, class) is the primary key (combination of columns with unique values) for this table.
+Each row of this table indicates the name of a student and the class in which they are enrolled.
+</pre>
+
+<p>&nbsp;</p>
+
+<p>Write a solution to find all the classes that have <strong>at least five students</strong>.</p>
+
+<p>Return the result table in <strong>any order</strong>.</p>
+
+<p>The&nbsp;result format is in the following example.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> 
+Courses table:
++---------+----------+
+| student | class    |
++---------+----------+
+| A       | Math     |
+| B       | English  |
+| C       | Math     |
+| D       | Biology  |
+| E       | Math     |
+| F       | Computer |
+| G       | Math     |
+| H       | Math     |
+| I       | Math     |
++---------+----------+
+<strong>Output:</strong> 
++---------+
+| class   |
++---------+
+| Math    |
++---------+
+<strong>Explanation:</strong> 
+- Math has 6 students, so we include it.
+- English has 1 student, so we do not include it.
+- Biology has 1 student, so we do not include it.
+- Computer has 1 student, so we do not include it.
+</pre>
+
+<!-- description:end -->
+
+## Solutions
+
+<!-- solution:start -->
+
+### Solution 1
+
+<!-- tabs:start -->
+
+#### MySQL
+
+```sql
+
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- problem:end -->

solution/0700-0799/0799.Champagne Tower/README_EN.md

@@ -27,51 +27,35 @@ tags:
 <p>Now after pouring some non-negative integer cups of champagne, return how full the <code>j<sup>th</sup></code> glass in the <code>i<sup>th</sup></code> row is (both <code>i</code> and <code>j</code> are 0-indexed.)</p>
 
 <p>&nbsp;</p>
-
 <p><strong class="example">Example 1:</strong></p>
 
 <pre>
-
 <strong>Input:</strong> poured = 1, query_row = 1, query_glass = 1
-
 <strong>Output:</strong> 0.00000
-
 <strong>Explanation:</strong> We poured 1 cup of champange to the top glass of the tower (which is indexed as (0, 0)). There will be no excess liquid so all the glasses under the top glass will remain empty.
-
 </pre>
 
 <p><strong class="example">Example 2:</strong></p>
 
 <pre>
-
 <strong>Input:</strong> poured = 2, query_row = 1, query_glass = 1
-
 <strong>Output:</strong> 0.50000
-
 <strong>Explanation:</strong> We poured 2 cups of champange to the top glass of the tower (which is indexed as (0, 0)). There is one cup of excess liquid. The glass indexed as (1, 0) and the glass indexed as (1, 1) will share the excess liquid equally, and each will get half cup of champange.
-
 </pre>
 
 <p><strong class="example">Example 3:</strong></p>
 
 <pre>
-
 <strong>Input:</strong> poured = 100000009, query_row = 33, query_glass = 17
-
 <strong>Output:</strong> 1.00000
-
 </pre>
 
 <p>&nbsp;</p>
-
 <p><strong>Constraints:</strong></p>
 
 <ul>
-
-    <li><code>0 &lt;=&nbsp;poured &lt;= 10<sup>9</sup></code></li>
-
-    <li><code>0 &lt;= query_glass &lt;= query_row&nbsp;&lt; 100</code></li>
-
+	<li><code>0 &lt;=&nbsp;poured &lt;= 10<sup>9</sup></code></li>
+	<li><code>0 &lt;= query_glass &lt;= query_row&nbsp;&lt; 100</code></li>
 </ul>
 
 <!-- description:end -->

solution/1000-1099/1016.Binary String With Substrings Representing 1 To N/README.md

@@ -5,7 +5,10 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/1000-1099/1016.Bi
 rating: 1779
 source: 第 129 场周赛 Q4
 tags:
+    - 位运算
+    - 哈希表
     - 字符串
+    - 滑动窗口
 ---
 
 <!-- problem:start -->

solution/1000-1099/1016.Binary String With Substrings Representing 1 To N/README_EN.md

@@ -5,7 +5,10 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/1000-1099/1016.Bi
 rating: 1779
 source: Weekly Contest 129 Q4
 tags:
+    - Bit Manipulation
+    - Hash Table
     - String
+    - Sliding Window
 ---
 
 <!-- problem:start -->