feat: update sql solution to lc problem: No.0615,0618 (doocs#1193)

* No.0615.Average Salary Departments VS Company
* No.0618.Students Report By Geography

Author: yanglbme

solution/0600-0699/0615.Average Salary Departments VS Company/README.md

@@ -105,4 +105,35 @@ SELECT DISTINCT
 FROM t;
 ```
 
+```sql
+# Write your MySQL query statement below
+WITH
+    S AS (
+        SELECT *
+        FROM
+            Salary
+            JOIN Employee USING (employee_id)
+    ),
+    T AS (
+        SELECT
+            date_format(pay_date, '%Y-%m') AS pay_month,
+            department_id,
+            avg(amount) OVER (
+                PARTITION BY pay_date, department_id
+            ) AS department_avg,
+            avg(amount) OVER (PARTITION BY pay_date) AS company_avg
+        FROM S
+    )
+SELECT
+    pay_month,
+    department_id,
+    CASE
+        WHEN avg(department_avg) > avg(company_avg) THEN 'higher'
+        WHEN avg(department_avg) < avg(company_avg) THEN 'lower'
+        ELSE 'same'
+    END AS comparison
+FROM T
+GROUP BY 1, 2;
+```
+
 <!-- tabs:end -->

solution/0600-0699/0615.Average Salary Departments VS Company/README_EN.md

@@ -116,4 +116,35 @@ SELECT DISTINCT
 FROM t;
 ```
 
+```sql
+# Write your MySQL query statement below
+WITH
+    S AS (
+        SELECT *
+        FROM
+            Salary
+            JOIN Employee USING (employee_id)
+    ),
+    T AS (
+        SELECT
+            date_format(pay_date, '%Y-%m') AS pay_month,
+            department_id,
+            avg(amount) OVER (
+                PARTITION BY pay_date, department_id
+            ) AS department_avg,
+            avg(amount) OVER (PARTITION BY pay_date) AS company_avg
+        FROM S
+    )
+SELECT
+    pay_month,
+    department_id,
+    CASE
+        WHEN avg(department_avg) > avg(company_avg) THEN 'higher'
+        WHEN avg(department_avg) < avg(company_avg) THEN 'lower'
+        ELSE 'same'
+    END AS comparison
+FROM T
+GROUP BY 1, 2;
+```
+
 <!-- tabs:end -->

solution/0600-0699/0615.Average Salary Departments VS Company/Solution.sql

@@ -1,23 +1,28 @@
 # Write your MySQL query statement below
 WITH
-    t AS (
+    S AS (
+        SELECT *
+        FROM
+            Salary
+            JOIN Employee USING (employee_id)
+    ),
+    T AS (
         SELECT
             date_format(pay_date, '%Y-%m') AS pay_month,
             department_id,
-            avg(amount) OVER (PARTITION BY pay_date) AS company_avg_amount,
             avg(amount) OVER (
                 PARTITION BY pay_date, department_id
-            ) AS department_avg_amount
-        FROM
-            Salary AS s
-            JOIN Employee AS e ON s.employee_id = e.employee_id
+            ) AS department_avg,
+            avg(amount) OVER (PARTITION BY pay_date) AS company_avg
+        FROM S
     )
-SELECT DISTINCT
+SELECT
     pay_month,
     department_id,
     CASE
-        WHEN company_avg_amount = department_avg_amount THEN 'same'
-        WHEN company_avg_amount < department_avg_amount THEN 'higher'
-        ELSE 'lower'
+        WHEN avg(department_avg) > avg(company_avg) THEN 'higher'
+        WHEN avg(department_avg) < avg(company_avg) THEN 'lower'
+        ELSE 'same'
     END AS comparison
-FROM t;
+FROM T
+GROUP BY 1, 2;

solution/0600-0699/0618.Students Report By Geography/README.md

@@ -60,31 +60,32 @@ Student table:
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：窗口函数 + GROUP BY**
+
+我们可以使用窗口函数 `row_number()` 来为每个大洲的学生编号，然后使用 `GROUP BY` 来将同一编号的学生聚合到一行中。
+
 <!-- tabs:start -->
 
 ### **SQL**
 
 ```sql
 # Write your MySQL query statement below
-with t as (
-    select
-        *,
-        row_number() over(
-            partition by continent
-            order by
-                name
-        ) rn
-    from
-        Student
-)
-select
-    max(if(continent = 'America', name, null)) America,
-    max(if(continent = 'Asia', name, null)) Asia,
-    max(if(continent = 'Europe', name, null)) Europe
-from
-    t
-group by
-    rn
+WITH
+    T AS (
+        SELECT
+            *,
+            row_number() OVER (
+                PARTITION BY continent
+                ORDER BY name
+            ) AS rk
+        FROM Student
+    )
+SELECT
+    max(if(continent = 'America', name, NULL)) AS 'America',
+    max(if(continent = 'Asia', name, NULL)) AS 'Asia',
+    max(if(continent = 'Europe', name, NULL)) AS 'Europe'
+FROM T
+GROUP BY rk;
 ```
 
 <!-- tabs:end -->

solution/0600-0699/0618.Students Report By Geography/README_EN.md

@@ -61,25 +61,22 @@ Student table:
 
 ```sql
 # Write your MySQL query statement below
-with t as (
-    select
-        *,
-        row_number() over(
-            partition by continent
-            order by
-                name
-        ) rn
-    from
-        Student
-)
-select
-    max(if(continent = 'America', name, null)) America,
-    max(if(continent = 'Asia', name, null)) Asia,
-    max(if(continent = 'Europe', name, null)) Europe
-from
-    t
-group by
-    rn
+WITH
+    T AS (
+        SELECT
+            *,
+            row_number() OVER (
+                PARTITION BY continent
+                ORDER BY name
+            ) AS rk
+        FROM Student
+    )
+SELECT
+    max(if(continent = 'America', name, NULL)) AS 'America',
+    max(if(continent = 'Asia', name, NULL)) AS 'Asia',
+    max(if(continent = 'Europe', name, NULL)) AS 'Europe'
+FROM T
+GROUP BY rk;
 ```
 
 <!-- tabs:end -->

solution/0600-0699/0618.Students Report By Geography/Solution.sql

@@ -1,17 +1,17 @@
 # Write your MySQL query statement below
 WITH
-    t AS (
+    T AS (
         SELECT
             *,
             row_number() OVER (
                 PARTITION BY continent
                 ORDER BY name
-            ) AS rn
+            ) AS rk
         FROM Student
     )
 SELECT
-    max(if(continent = 'America', name, NULL)) AS America,
-    max(if(continent = 'Asia', name, NULL)) AS Asia,
-    max(if(continent = 'Europe', name, NULL)) AS Europe
-FROM t
-GROUP BY rn;
+    max(if(continent = 'America', name, NULL)) AS 'America',
+    max(if(continent = 'Asia', name, NULL)) AS 'Asia',
+    max(if(continent = 'Europe', name, NULL)) AS 'Europe'
+FROM T
+GROUP BY rk;