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Commit 102e2c3

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feat: update solutions to lc problems: No.0534,0608,1264,1965 (#1791)
* No.0534.Game Play Analysis III * No.0608.Tree Node * No.1264.Page Recommendations * No.1965.Employees With Missing Information
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‎solution/0500-0599/0534.Game Play Analysis III/README.md‎

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<!-- 这里可写通用的实现逻辑 -->
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**方法一:SUM() OVER() 窗口函数**
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**方法一:使用窗口函数**
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我们可以使用 `SUM() OVER()` 窗口函数来计算每个玩家到目前为止玩了多少游戏。在 `OVER()` 子句中,我们使用 `PARTITION BY` 子句将玩家分组,然后使用 `ORDER BY` 子句按日期排序。
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我们可以使用窗口函数 `SUM() OVER()`,按照 `player_id` 分组,按照 `event_date` 排序,计算每个用户截止到当前日期的游戏总数。
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**方法二:使用自连接 + 分组**
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我们也可以使用自连接,将 `Activity` 表自连接,连接条件为 `t1.player_id = t2.player_id AND t1.event_date >= t2.event_date`,然后按照 `t1.player_id` 和 `t1.event_date` 分组,累计 `t2.games_played`,得到每个用户截止到当前日期的游戏总数。
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<!-- tabs:start -->
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FROM Activity;
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```
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```sql
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# Write your MySQL query statement below
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SELECT
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t1.player_id,
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t1.event_date,
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sum(t2.games_played) AS games_played_so_far
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FROM
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Activity AS t1,
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Activity AS t2
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WHERE t1.player_id = t2.player_id AND t1.event_date >= t2.event_date
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GROUP BY 1, 2;
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```
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```sql
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# Write your MySQL query statement below
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SELECT
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t1.player_id,
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t1.event_date,
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sum(t2.games_played) AS games_played_so_far
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FROM
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Activity AS t1
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CROSS JOIN Activity AS t2 ON t1.player_id = t2.player_id AND t1.event_date >= t2.event_date
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GROUP BY 1, 2;
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```
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‎solution/0500-0599/0534.Game Play Analysis III/README_EN.md‎

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## Solutions
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**Solution 1: Window Function**
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We can use the window function `SUM() OVER()` to group by `player_id`, sort by `event_date`, and calculate the total number of games played by each user up to the current date.
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**Solution 2: Self-Join + Group By**
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We can also use a self-join to join the `Activity` table with itself on the condition of `t1.player_id = t2.player_id AND t1.event_date >= t2.event_date`, and then group by `t1.player_id` and `t1.event_date`, and calculate the cumulative sum of `t2.games_played`. This will give us the total number of games played by each user up to the current date.
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### **SQL**
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FROM Activity;
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```
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```sql
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# Write your MySQL query statement below
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SELECT
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t1.player_id,
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t1.event_date,
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sum(t2.games_played) AS games_played_so_far
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FROM
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Activity AS t1,
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Activity AS t2
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WHERE t1.player_id = t2.player_id AND t1.event_date >= t2.event_date
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GROUP BY 1, 2;
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```
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```sql
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# Write your MySQL query statement below
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SELECT
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t1.player_id,
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t1.event_date,
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sum(t2.games_played) AS games_played_so_far
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FROM
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Activity AS t1
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CROSS JOIN Activity AS t2 ON t1.player_id = t2.player_id AND t1.event_date >= t2.event_date
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GROUP BY 1, 2;
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```
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‎solution/0600-0699/0608.Tree Node/README.md‎

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<!-- 这里可写通用的实现逻辑 -->
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**方法一:条件判断 + 子查询**
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我们可以使用 `CASE WHEN` 条件判断语句来判断每个节点的类型,具体地:
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- 如果一个节点的 `p_id` 为 `NULL`,则该节点为根节点;
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- 否则,如果一个节点是另一个节点的父节点(这里我们使用子查询来判断),则该节点为内部节点;
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- 否则,该节点为叶子节点。
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### **SQL**

‎solution/0600-0699/0608.Tree Node/README_EN.md‎

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## Solutions
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**Solution 1: Conditional Statements + Subquery**
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We can use the `CASE WHEN` conditional statement to determine the type of each node as follows:
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- If a node's `p_id` is `NULL`, then it is a root node.
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- Otherwise, if a node is the parent node of another node (we use a subquery to determine this), then it is an internal node.
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- Otherwise, it is a leaf node.
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### **SQL**

‎solution/1200-1299/1264.Page Recommendations/README.md‎

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<!-- 这里可写通用的实现逻辑 -->
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**方法一:合并 + 等值连接 + 子查询**
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我们先查出所有与 `user_id = 1` 的用户是朋友的用户,记录在 `T` 表中,然后再查出所有在 `T` 表中的用户喜欢的页面,最后排除掉 `user_id = 1` 喜欢的页面即可。
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### **SQL**
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```sql
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# Write your MySQL query statement below
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WITH
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T AS (
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SELECT user1_id AS user_id FROM Friendship WHERE user2_id = 1
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UNION
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SELECT user2_id AS user_id FROM Friendship WHERE user1_id = 1
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)
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SELECT DISTINCT page_id AS recommended_page
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FROM
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T
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JOIN Likes USING (user_id)
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WHERE page_id NOT IN (SELECT page_id FROM Likes WHERE user_id = 1);
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```
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```sql
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# Write your MySQL query statement below
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SELECT DISTINCT page_id AS recommended_page

‎solution/1200-1299/1264.Page Recommendations/README_EN.md‎

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## Solutions
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**Solution 1: Union + Equi-Join + Subquery**
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First, we query all users who are friends with `user_id = 1` and record them in the `T` table. Then, we query all pages that users in the `T` table like, and finally exclude the pages that `user_id = 1` likes.
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### **SQL**
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```sql
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# Write your MySQL query statement below
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WITH
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T AS (
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SELECT user1_id AS user_id FROM Friendship WHERE user2_id = 1
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UNION
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SELECT user2_id AS user_id FROM Friendship WHERE user1_id = 1
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)
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SELECT DISTINCT page_id AS recommended_page
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FROM
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T
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JOIN Likes USING (user_id)
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WHERE page_id NOT IN (SELECT page_id FROM Likes WHERE user_id = 1);
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```
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```sql
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SELECT DISTINCT page_id AS recommended_page

‎solution/1900-1999/1965.Employees With Missing Information/README.md‎

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<!-- 这里可写通用的实现逻辑 -->
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**方法一:子查询 + 合并**
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我们可以先从 `Employees` 表中找出所有不在 `Salaries` 表中的 `employee_id`,再从 `Salaries` 表中找出所有不在 `Employees` 表中的 `employee_id`,最后将两个结果合并,然后按照 `employee_id` 排序即可。
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### **SQL**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```sql
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# Write your MySQL query statement below
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SELECT employee_id
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FROM Employees AS e
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WHERE
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e.employee_id NOT IN (
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SELECT employee_id
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FROM Salaries
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)
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FROM Employees
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WHERE employee_id NOT IN (SELECT employee_id FROM Salaries)
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UNION
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SELECT employee_id
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FROM Salaries AS s
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WHERE
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s.employee_id NOT IN (
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SELECT employee_id
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FROM Employees
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)
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ORDER BY employee_id;
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FROM Salaries
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WHERE employee_id NOT IN (SELECT employee_id FROM Employees)
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ORDER BY 1;
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```
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‎solution/1900-1999/1965.Employees With Missing Information/README_EN.md‎

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## Solutions
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**Solution 1: Subquery + Union**
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We can first find all `employee_id` that are not in the `Salaries` table from the `Employees` table, and then find all `employee_id` that are not in the `Employees` table from the `Salaries` table. Finally, we can combine the two results using the `UNION` operator, and sort the result by `employee_id`.
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### **SQL**
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```sql
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# Write your MySQL query statement below
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SELECT employee_id
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FROM Employees AS e
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WHERE
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e.employee_id NOT IN (
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SELECT employee_id
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FROM Salaries
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)
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FROM Employees
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WHERE employee_id NOT IN (SELECT employee_id FROM Salaries)
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UNION
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SELECT employee_id
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FROM Salaries AS s
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WHERE
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s.employee_id NOT IN (
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SELECT employee_id
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FROM Employees
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)
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ORDER BY employee_id;
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FROM Salaries
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WHERE employee_id NOT IN (SELECT employee_id FROM Employees)
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ORDER BY 1;
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```
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<!-- tabs:end -->
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# Write your MySQL query statement below
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SELECT employee_id
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FROM Employees AS e
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WHERE
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e.employee_id NOT IN (
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SELECT employee_id
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FROM Salaries
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)
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FROM Employees
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WHERE employee_id NOT IN (SELECT employee_id FROM Salaries)
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UNION
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SELECT employee_id
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FROM Salaries AS s
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WHERE
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s.employee_id NOT IN (
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SELECT employee_id
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FROM Employees
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)
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ORDER BY employee_id;
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FROM Salaries
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WHERE employee_id NOT IN (SELECT employee_id FROM Employees)
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ORDER BY 1;

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