chore: update lc problems (#1455)

Author: yanglbme

solution/0000-0099/0055.Jump Game/README.md

@@ -6,37 +6,35 @@
 
 <!-- 这里写题目描述 -->
 
-<p>给定一个非负整数数组 <code>nums</code> ，你最初位于数组的 <strong>第一个下标</strong> 。</p>
+<p>给你一个非负整数数组&nbsp;<code>nums</code> ，你最初位于数组的 <strong>第一个下标</strong> 。数组中的每个元素代表你在该位置可以跳跃的最大长度。</p>
 
-<p>数组中的每个元素代表你在该位置可以跳跃的最大长度。</p>
+<p>判断你是否能够到达最后一个下标，如果可以，返回 <code>true</code> ；否则，返回 <code>false</code> 。</p>
 
-<p>判断你是否能够到达最后一个下标。</p>
+<p>&nbsp;</p>
 
-<p> </p>
-
-<p><strong>示例 1：</strong></p>
+<p><strong>示例&nbsp;1：</strong></p>
 
 <pre>
 <strong>输入：</strong>nums = [2,3,1,1,4]
 <strong>输出：</strong>true
 <strong>解释：</strong>可以先跳 1 步，从下标 0 到达下标 1, 然后再从下标 1 跳 3 步到达最后一个下标。
 </pre>
 
-<p><strong>示例 2：</strong></p>
+<p><strong>示例&nbsp;2：</strong></p>
 
 <pre>
 <strong>输入：</strong>nums = [3,2,1,0,4]
 <strong>输出：</strong>false
 <strong>解释：</strong>无论怎样，总会到达下标为 3 的位置。但该下标的最大跳跃长度是 0 ， 所以永远不可能到达最后一个下标。
 </pre>
 
-<p> </p>
+<p>&nbsp;</p>
 
 <p><strong>提示：</strong></p>
 
 <ul>
-	<li><code>1 <= nums.length <= 3 * 10<sup>4</sup></code></li>
-	<li><code>0 <= nums[i] <= 10<sup>5</sup></code></li>
+	<li><code>1 &lt;= nums.length &lt;= 10<sup>4</sup></code></li>
+	<li><code>0 &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>
 </ul>
 
 ## 解法

solution/0100-0199/0175.Combine Two Tables/README_EN.md

@@ -14,7 +14,7 @@
 | lastName    | varchar |
 | firstName   | varchar |
 +-------------+---------+
-personId is the primary key column for this table.
+personId is the primary key (column with unique values) for this table.
 This table contains information about the ID of some persons and their first and last names.
 </pre>
 
@@ -31,17 +31,17 @@ This table contains information about the ID of some persons and their first and
 | city        | varchar |
 | state       | varchar |
 +-------------+---------+
-addressId is the primary key column for this table.
+addressId is the primary key (column with unique values) for this table.
 Each row of this table contains information about the city and state of one person with ID = PersonId.
 </pre>
 
 <p>&nbsp;</p>
 
-<p>Write an SQL query to report the first name, last name, city, and state of each person in the <code>Person</code> table. If the address of a <code>personId</code> is not present in the <code>Address</code> table, report <code>null</code> instead.</p>
+<p>Write a solution to report the first name, last name, city, and state of each person in the <code>Person</code> table. If the address of a <code>personId</code> is not present in the <code>Address</code> table, report <code>null</code> instead.</p>
 
 <p>Return the result table in <strong>any order</strong>.</p>
 
-<p>The query 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/0100-0199/0181.Employees Earning More Than Their Managers/README_EN.md

@@ -15,17 +15,17 @@
 | salary      | int     |
 | managerId   | int     |
 +-------------+---------+
-id is the primary key column for this table.
+id is the primary key (column with unique values) for this table.
 Each row of this table indicates the ID of an employee, their name, salary, and the ID of their manager.
 </pre>
 
 <p>&nbsp;</p>
 
-<p>Write an SQL query to find the employees who earn more than their managers.</p>
+<p>Write a solution&nbsp;to find the employees who earn more than their managers.</p>
 
 <p>Return the result table in <strong>any order</strong>.</p>
 
-<p>The query 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/0100-0199/0182.Duplicate Emails/README_EN.md

@@ -13,17 +13,17 @@
 | id          | int     |
 | email       | varchar |
 +-------------+---------+
-id is the primary key column for this table.
+id is the primary key (column with unique values) for this table.
 Each row of this table contains an email. The emails will not contain uppercase letters.
 </pre>
 
 <p>&nbsp;</p>
 
-<p>Write an SQL query to report all the duplicate emails. Note that it&#39;s guaranteed that the email&nbsp;field is not NULL.</p>
+<p>Write a solution to report all the duplicate emails. Note that it&#39;s guaranteed that the email&nbsp;field is not NULL.</p>
 
 <p>Return the result table in <strong>any order</strong>.</p>
 
-<p>The query result format is in the following example.</p>
+<p>The&nbsp;result format is in the following example.</p>
 
 <p>&nbsp;</p>
 <p><strong class="example">Example 1:</strong></p>

solution/0100-0199/0185.Department Top Three Salaries/README.md

@@ -17,8 +17,8 @@
 | salary       | int     |
 | departmentId | int     |
 +--------------+---------+
-Id是该表的主键列。
-departmentId是Department表中ID的外键。
+id 是该表的主键列(具有唯一值的列)。
+departmentId 是 Department 表中 ID 的外键（reference 列）。
 该表的每一行都表示员工的ID、姓名和工资。它还包含了他们部门的ID。
 </pre>
 
@@ -33,19 +33,19 @@ departmentId是Department表中ID的外键。
 | id          | int     |
 | name        | varchar |
 +-------------+---------+
-Id是该表的主键列。
+id 是该表的主键列(具有唯一值的列)。
 该表的每一行表示部门ID和部门名。
 </pre>
 
 <p>&nbsp;</p>
 
 <p>公司的主管们感兴趣的是公司每个部门中谁赚的钱最多。一个部门的 <strong>高收入者</strong> 是指一个员工的工资在该部门的 <strong>不同</strong> 工资中 <strong>排名前三</strong> 。</p>
 
-<p>编写一个SQL查询，找出每个部门中 <strong>收入高的员工</strong> 。</p>
+<p>编写解决方案，找出每个部门中 <strong>收入高的员工</strong> 。</p>
 
 <p>以 <strong>任意顺序</strong> 返回结果表。</p>
 
-<p>查询结果格式如下所示。</p>
+<p>返回结果格式如下所示。</p>
 
 <p>&nbsp;</p>
 

solution/0100-0199/0185.Department Top Three Salaries/README_EN.md

@@ -15,8 +15,8 @@
 | salary       | int     |
 | departmentId | int     |
 +--------------+---------+
-id is the primary key column for this table.
-departmentId is a foreign key of the ID from the <code>Department </code>table.
+id is the primary key (column with unique values) for this table.
+departmentId is a foreign key (reference column) of the ID from the <code>Department </code>table.
 Each row of this table indicates the ID, name, and salary of an employee. It also contains the ID of their department.
 </pre>
 
@@ -31,19 +31,19 @@ Each row of this table indicates the ID, name, and salary of an employee. It als
 | id          | int     |
 | name        | varchar |
 +-------------+---------+
-id is the primary key column for this table.
+id is the primary key (column with unique values) for this table.
 Each row of this table indicates the ID of a department and its name.
 </pre>
 
 <p>&nbsp;</p>
 
 <p>A company&#39;s executives are interested in seeing who earns the most money in each of the company&#39;s departments. A <strong>high earner</strong> in a department is an employee who has a salary in the <strong>top three unique</strong> salaries for that department.</p>
 
-<p>Write an SQL query to find the employees who are <strong>high earners</strong> in each of the departments.</p>
+<p>Write a solution to find the employees who are <strong>high earners</strong> in each of the departments.</p>
 
 <p>Return the result table <strong>in any order</strong>.</p>
 
-<p>The query result format is in the following example.</p>
+<p>The&nbsp;result format is in the following example.</p>
 
 <p>&nbsp;</p>
 <p><strong class="example">Example 1:</strong></p>

solution/0100-0199/0196.Delete Duplicate Emails/README.md

@@ -15,13 +15,13 @@
 | id          | int     |
 | email       | varchar |
 +-------------+---------+
-在 SQL 中，id 是该表的主键列。
+id 是该表的主键列(具有唯一值的列)。
 该表的每一行包含一封电子邮件。电子邮件将不包含大写字母。
 </pre>
 
 <p>&nbsp;</p>
 
-<p><strong>删除</strong> 所有重复的电子邮件，只保留一个具有最小 <code>id</code> 的唯一电子邮件。</p>
+<p>编写解决方案<strong> 删除</strong> 所有重复的电子邮件，只保留一个具有最小 <code>id</code> 的唯一电子邮件。</p>
 
 <p>（对于 SQL 用户，请注意你应该编写一个 <code>DELETE</code> 语句而不是 <code>SELECT</code> 语句。）</p>
 

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

@@ -21,7 +21,7 @@
 | status      | enum     |
 | request_at  | date     |     
 +-------------+----------+
-id 是这张表的主键。
+id 是这张表的主键（具有唯一值的列）。
 这张表中存所有出租车的行程信息。每段行程有唯一 id ，其中 client_id 和 driver_id 是 Users 表中 users_id 的外键。
 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>
@@ -42,7 +42,7 @@ status 是一个表示行程状态的枚举类型，枚举成员为(‘completed
 | banned      | enum     |
 | role        | enum     |
 +-------------+----------+
-users_id 是这张表的主键。
+users_id 是这张表的主键（具有唯一值的列）。
 这张表中存所有用户，每个用户都有一个唯一的 users_id ，role 是一个表示用户身份的枚举类型，枚举成员为 (‘client’, ‘driver’, ‘partner’) 。
 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>
@@ -51,15 +51,15 @@ banned 是一个表示用户是否被禁止的枚举类型，枚举成员为 (
 
 <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>写一段 SQL 语句查出&nbsp;<code>"2013-10-01"</code><strong>&nbsp;</strong>至&nbsp;<code>"2013-10-03"</code><strong>&nbsp;</strong>期间非禁止用户（<strong>乘客和司机都必须未被禁止</strong>）的取消率。非禁止用户即 banned 为 No 的用户，禁止用户即 banned 为 Yes 的用户。</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>期间非禁止用户（<strong>乘客和司机都必须未被禁止</strong>）的取消率。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>
 
-<p>返回结果表中的数据可以按任意顺序组织。其中取消率 <code>Cancellation Rate</code> 需要四舍五入保留 <strong>两位小数</strong> 。</p>
+<p>返回结果表中的数据 <strong>无顺序要求</strong> 。</p>
 
-<p>查询结果格式如下例所示。</p>
+<p>结果格式如下例所示。</p>
 
 <p>&nbsp;</p>
 
-<p><strong>示例：</strong></p>
+<p><strong>示例 1：</strong></p>
 
 <pre>
 <strong>输入：</strong> 
@@ -78,7 +78,6 @@ Trips 表：
 | 9  | 3         | 10        | 12      | completed           | 2013-10-03 |
 | 10 | 4         | 13        | 12      | cancelled_by_driver | 2013-10-03 |
 +----+-----------+-----------+---------+---------------------+------------+
-
 Users 表：
 +----------+--------+--------+
 | users_id | banned | role   |

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

@@ -17,9 +17,9 @@
 | status      | enum     |
 | request_at  | date     |     
 +-------------+----------+
-id is the primary key for this table.
+id is the primary key (column with unique values) for this table.
 The table holds all taxi trips. Each trip has a unique id, while client_id and driver_id are foreign keys to the users_id at the Users table.
-Status is an ENUM type of ('completed', 'cancelled_by_driver', 'cancelled_by_client').
+Status is an ENUM (category) type of ('completed', 'cancelled_by_driver', 'cancelled_by_client').
 </pre>
 
 <p>&nbsp;</p>
@@ -34,20 +34,20 @@ Status is an ENUM type of (&#39;completed&#39;, &#39;cancelled_by_driver&#39;, &
 | banned      | enum     |
 | role        | enum     |
 +-------------+----------+
-users_id is the primary key for this table.
+users_id is the primary key (column with unique values) for this table.
 The table holds all users. Each user has a unique users_id, and role is an ENUM type of (&#39;client&#39;, &#39;driver&#39;, &#39;partner&#39;).
-banned is an ENUM type of (&#39;Yes&#39;, &#39;No&#39;).
+banned is an ENUM (category) type of (&#39;Yes&#39;, &#39;No&#39;).
 </pre>
 
 <p>&nbsp;</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 SQL query 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>. Round <code>Cancellation Rate</code> to <strong>two decimal</strong> points.</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>. 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 query result format is in the following example.</p>
+<p>The&nbsp;result format is in the following example.</p>
 
 <p>&nbsp;</p>
 <p><strong class="example">Example 1:</strong></p>

solution/0400-0499/0427.Construct Quad Tree/README.md

@@ -8,18 +8,17 @@
 
 <p>给你一个 <code>n * n</code> 矩阵 <code>grid</code> ，矩阵由若干 <code>0</code> 和 <code>1</code> 组成。请你用四叉树表示该矩阵 <code>grid</code> 。</p>
 
-<p>你需要返回能表示矩阵的 四叉树 的根结点。</p>
-
-<p>注意，当 <code>isLeaf</code> 为 <strong>False </strong>时，你可以把 <strong>True</strong> 或者 <strong>False</strong> 赋值给节点，两种值都会被判题机制 <strong>接受</strong> 。</p>
+<p>你需要返回能表示矩阵 <code>grid</code> 的 四叉树 的根结点。</p>
 
 <p>四叉树数据结构中，每个内部节点只有四个子节点。此外，每个节点都有两个属性：</p>
 
 <ul>
-	<li><code>val</code>：储存叶子结点所代表的区域的值。1 对应 <strong>True</strong>，0 对应 <strong>False</strong>；</li>
+	<li><code>val</code>：储存叶子结点所代表的区域的值。1 对应 <strong>True</strong>，0 对应 <strong>False</strong>。注意，当 <code>isLeaf</code> 为 <strong>False </strong>时，你可以把 <strong>True</strong> 或者 <strong>False</strong> 赋值给节点，两种值都会被判题机制 <strong>接受</strong> 。</li>
 	<li><code>isLeaf</code>: 当这个节点是一个叶子结点时为 <strong>True</strong>，如果它有 4 个子节点则为 <strong>False</strong> 。</li>
 </ul>
 
-<pre>class Node {
+<pre>
+class Node {
     public boolean val;
 &nbsp; &nbsp; public boolean isLeaf;
 &nbsp; &nbsp; public Node topLeft;
@@ -36,13 +35,13 @@
 	<li>使用适当的子网格递归每个子节点。</li>
 </ol>
 
-<p><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0400-0499/0427.Construct%20Quad%20Tree/images/new_top.png" style="height: 181px; width: 777px;"></p>
+<p><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0400-0499/0427.Construct%20Quad%20Tree/images/new_top.png" style="height: 181px; width: 777px;" /></p>
 
 <p>如果你想了解更多关于四叉树的内容，可以参考 <a href="https://en.wikipedia.org/wiki/Quadtree">wiki</a> 。</p>
 
 <p><strong>四叉树格式：</strong></p>
 
-<p>输出为使用层序遍历后四叉树的序列化形式，其中 <code>null</code> 表示路径终止符，其下面不存在节点。</p>
+<p>你不需要阅读本节来解决这个问题。只有当你想了解输出格式时才会这样做。输出为使用层序遍历后四叉树的序列化形式，其中 <code>null</code> 表示路径终止符，其下面不存在节点。</p>
 
 <p>它与二叉树的序列化非常相似。唯一的区别是节点以列表形式表示 <code>[isLeaf, val]</code> 。</p>
 
@@ -52,44 +51,28 @@
 
 <p><strong>示例 1：</strong></p>
 
-<p><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0400-0499/0427.Construct%20Quad%20Tree/images/grid1.png" style="height: 99px; width: 777px;"></p>
+<p><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0400-0499/0427.Construct%20Quad%20Tree/images/grid1.png" style="height: 99px; width: 777px;" /></p>
 
-<pre><strong>输入：</strong>grid = [[0,1],[1,0]]
+<pre>
+<strong>输入：</strong>grid = [[0,1],[1,0]]
 <strong>输出：</strong>[[0,1],[1,0],[1,1],[1,1],[1,0]]
 <strong>解释：</strong>此示例的解释如下：
 请注意，在下面四叉树的图示中，0 表示 false，1 表示 True 。
-<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0400-0499/0427.Construct%20Quad%20Tree/images/e1tree.png" style="height: 186px; width: 777px;">
+<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0400-0499/0427.Construct%20Quad%20Tree/images/e1tree.png" style="height: 186px; width: 777px;" />
 </pre>
 
 <p><strong>示例 2：</strong></p>
 
-<p><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0400-0499/0427.Construct%20Quad%20Tree/images/e2mat.png" style="height: 343px; width: 777px;"></p>
+<p><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0400-0499/0427.Construct%20Quad%20Tree/images/e2mat.png" style="height: 343px; width: 777px;" /></p>
 
-<pre><strong>输入：</strong>grid = [[1,1,1,1,0,0,0,0],[1,1,1,1,0,0,0,0],[1,1,1,1,1,1,1,1],[1,1,1,1,1,1,1,1],[1,1,1,1,0,0,0,0],[1,1,1,1,0,0,0,0],[1,1,1,1,0,0,0,0],[1,1,1,1,0,0,0,0]]
+<pre>
+<strong>输入：</strong>grid = [[1,1,1,1,0,0,0,0],[1,1,1,1,0,0,0,0],[1,1,1,1,1,1,1,1],[1,1,1,1,1,1,1,1],[1,1,1,1,0,0,0,0],[1,1,1,1,0,0,0,0],[1,1,1,1,0,0,0,0],[1,1,1,1,0,0,0,0]]
 <strong>输出：</strong>[[0,1],[1,1],[0,1],[1,1],[1,0],null,null,null,null,[1,0],[1,0],[1,1],[1,1]]
 <strong>解释：</strong>网格中的所有值都不相同。我们将网格划分为四个子网格。
 topLeft，bottomLeft 和 bottomRight 均具有相同的值。
 topRight 具有不同的值，因此我们将其再分为 4 个子网格，这样每个子网格都具有相同的值。
 解释如下图所示：
-<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0400-0499/0427.Construct%20Quad%20Tree/images/e2tree.png" style="height: 328px; width: 777px;">
-</pre>
-
-<p><strong>示例 3：</strong></p>
-
-<pre><strong>输入：</strong>grid = [[1,1],[1,1]]
-<strong>输出：</strong>[[1,1]]
-</pre>
-
-<p><strong>示例 4：</strong></p>
-
-<pre><strong>输入：</strong>grid = [[0]]
-<strong>输出：</strong>[[1,0]]
-</pre>
-
-<p><strong>示例 5：</strong></p>
-
-<pre><strong>输入：</strong>grid = [[1,1,0,0],[1,1,0,0],[0,0,1,1],[0,0,1,1]]
-<strong>输出：</strong>[[0,1],[1,1],[1,0],[1,0],[1,1]]
+<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0400-0499/0427.Construct%20Quad%20Tree/images/e2tree.png" style="height: 328px; width: 777px;" />
 </pre>
 
 <p>&nbsp;</p>
@@ -98,7 +81,7 @@ topRight 具有不同的值，因此我们将其再分为 4 个子网格，这
 
 <ol>
 	<li><code>n == grid.length == grid[i].length</code></li>
-	<li><code>n == 2^x</code> 其中 <code>0 &lt;= x &lt;= 6</code></li>
+	<li><code>n == 2<sup>x</sup></code> 其中 <code>0 &lt;= x &lt;= 6</code></li>
 </ol>
 
 ## 解法