feat: update sql solutions to lc problems: No.0511,0512,0534

.github/workflows/black-lint.yml

@@ -1,8 +1,13 @@
 name: black-linter
 
 on:
-  pull_request:
-    branches: [main]
+  pull_request_target:
+    types:
+      - opened
+      - edited
+      - synchronize
+    branches:
+      - main
 
 jobs:
   build:

.github/workflows/clang-format-lint.yml

@@ -1,8 +1,13 @@
 name: clang-format-linter
 
 on:
-  pull_request:
-    branches: [main]
+  pull_request_target:
+    types:
+      - opened
+      - edited
+      - synchronize
+    branches:
+      - main
 
 jobs:
   build:

.github/workflows/pr-checker.yml

@@ -68,4 +68,5 @@ jobs:
             In addition, the coding style (such as the naming of variables and functions) should be as consistent as possible with the existing code in the project. 
 
             <!-- Sticky Pull Request Comment -->
+          emojis: '+1, laugh'
           body-include: '<!-- Sticky Pull Request Comment -->'

solution/0500-0599/0511.Game Play Analysis I/README.md

@@ -54,14 +54,17 @@ Result 表：
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：GROUP BY**
+
+我们可以用 `GROUP BY` 对 `player_id` 进行分组，然后取每一组中最小的 `event_date` 作为玩家第一次登录平台的日期。
+
 <!-- tabs:start -->
 
 ### **SQL**
 
 ```sql
-SELECT
-    player_id,
-    MIN(event_date) AS first_login
+# Write your MySQL query statement below
+SELECT player_id, min(event_date) AS first_login
 FROM Activity
 GROUP BY player_id;
 ```

solution/0500-0599/0511.Game Play Analysis I/README_EN.md

@@ -60,9 +60,8 @@ Activity table:
 ### **SQL**
 
 ```sql
-SELECT
-    player_id,
-    MIN(event_date) AS first_login
+# Write your MySQL query statement below
+SELECT player_id, min(event_date) AS first_login
 FROM Activity
 GROUP BY player_id;
 ```

solution/0500-0599/0511.Game Play Analysis I/Solution.sql

@@ -1,5 +1,4 @@
-SELECT
-    player_id,
-    MIN(event_date) AS first_login
+# Write your MySQL query statement below
+SELECT player_id, min(event_date) AS first_login
 FROM Activity
 GROUP BY player_id;

solution/0500-0599/0512.Game Play Analysis II/README.md

@@ -51,6 +51,14 @@ Result table:
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：子查询**
+
+我们可以使用 `GROUP BY` 和 `MIN` 函数来找到每个玩家的第一次登录日期，然后使用联合键子查询来找到每个玩家的第一次登录设备。
+
+**方法二：窗口函数**
+
+我们可以使用窗口函数 `rank()`，它可以为每个玩家的每个登录日期分配一个排名，然后我们可以选择排名为 $1$ 的行。
+
 <!-- tabs:start -->
 
 ### **SQL**
@@ -71,4 +79,21 @@ WHERE
     );
 ```
 
+```sql
+# Write your MySQL query statement below
+WITH
+    T AS (
+        SELECT
+            *,
+            rank() OVER (
+                PARTITION BY player_id
+                ORDER BY event_date
+            ) AS rk
+        FROM Activity
+    )
+SELECT player_id, device_id
+FROM T
+WHERE rk = 1;
+```
+
 <!-- tabs:end -->

solution/0500-0599/0512.Game Play Analysis II/README_EN.md

@@ -75,4 +75,21 @@ WHERE
     );
 ```
 
+```sql
+# Write your MySQL query statement below
+WITH
+    T AS (
+        SELECT
+            *,
+            rank() OVER (
+                PARTITION BY player_id
+                ORDER BY event_date
+            ) AS rk
+        FROM Activity
+    )
+SELECT player_id, device_id
+FROM T
+WHERE rk = 1;
+```
+
 <!-- tabs:end -->

solution/0500-0599/0534.Game Play Analysis III/README.md

@@ -59,20 +59,24 @@ Result table:
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：SUM() OVER() 窗口函数**
+
+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。
+
 <!-- tabs:start -->
 
 ### **SQL**
 
 ```sql
+# Write your MySQL query statement below
 SELECT
     player_id,
     event_date,
-    SUM(games_played) OVER (
+    sum(games_played) OVER (
         PARTITION BY player_id
         ORDER BY event_date
     ) AS games_played_so_far
-FROM Activity
-ORDER BY 1, 2;
+FROM Activity;
 ```
 
 <!-- tabs:end -->

solution/0500-0599/0534.Game Play Analysis III/README_EN.md

@@ -66,15 +66,15 @@ Note that for each player we only care about the days when the player logged in.
 ### **SQL**
 
 ```sql
+# Write your MySQL query statement below
 SELECT
     player_id,
     event_date,
-    SUM(games_played) OVER (
+    sum(games_played) OVER (
         PARTITION BY player_id
         ORDER BY event_date
     ) AS games_played_so_far
-FROM Activity
-ORDER BY 1, 2;
+FROM Activity;
 ```
 
 <!-- tabs:end -->

solution/0500-0599/0534.Game Play Analysis III/Solution.sql

@@ -1,9 +1,9 @@
+# Write your MySQL query statement below
 SELECT
     player_id,
     event_date,
-    SUM(games_played) OVER (
+    sum(games_played) OVER (
         PARTITION BY player_id
         ORDER BY event_date
     ) AS games_played_so_far
-FROM Activity
-ORDER BY 1, 2;
+FROM Activity;