feat: add sql solutions to lc problems: No.0601,0615 (doocs#1099)

* No.0601.Human Traffic of Stadium
* No.0615.Average Salary Departments VS Company

Author: yanglbme

solution/0600-0699/0601.Human Traffic of Stadium/README.md

@@ -68,26 +68,24 @@ id</strong> 为 5、6、7、8 的四行 id 连续，并且每行都有 &gt;= 100
 
 <!-- tabs:start -->
 
-### **Python3**
-
-<!-- 这里可写当前语言的特殊实现逻辑 -->
-
-```python
-
-```
-
-### **Java**
-
-<!-- 这里可写当前语言的特殊实现逻辑 -->
-
-```java
-
-```
-
-### **...**
-
-```
-
+### **SQL**
+
+```sql
+# Write your MySQL query statement below
+WITH
+    s AS (
+        SELECT *, id - row_number() OVER (ORDER BY id) AS rk
+        FROM Stadium
+        WHERE people >= 100
+    ),
+    t AS (
+        SELECT *, count(*) OVER (PARTITION BY rk) AS cnt
+        FROM s
+    )
+SELECT id, visit_date, people
+FROM t
+WHERE cnt >= 3
+ORDER BY visit_date;
 ```
 
 <!-- tabs:end -->

solution/0600-0699/0601.Human Traffic of Stadium/README_EN.md

@@ -63,22 +63,24 @@ The rows with ids 2 and 3 are not included because we need at least three consec
 
 <!-- tabs:start -->
 
-### **Python3**
-
-```python
-
-```
-
-### **Java**
-
-```java
-
-```
-
-### **...**
-
-```
-
+### **SQL**
+
+```sql
+# Write your MySQL query statement below
+WITH
+    s AS (
+        SELECT *, id - row_number() OVER (ORDER BY id) AS rk
+        FROM Stadium
+        WHERE people >= 100
+    ),
+    t AS (
+        SELECT *, count(*) OVER (PARTITION BY rk) AS cnt
+        FROM s
+    )
+SELECT id, visit_date, people
+FROM t
+WHERE cnt >= 3
+ORDER BY visit_date;
 ```
 
 <!-- tabs:end -->

solution/0600-0699/0601.Human Traffic of Stadium/Solution.sql

@@ -0,0 +1,15 @@
+# Write your MySQL query statement below
+WITH
+    s AS (
+        SELECT *, id - row_number() OVER (ORDER BY id) AS rk
+        FROM Stadium
+        WHERE people >= 100
+    ),
+    t AS (
+        SELECT *, count(*) OVER (PARTITION BY rk) AS cnt
+        FROM s
+    )
+SELECT id, visit_date, people
+FROM t
+WHERE cnt >= 3
+ORDER BY visit_date;

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

@@ -79,8 +79,30 @@
 
 ### **SQL**
 
-```
-
+```sql
+# Write your MySQL query statement below
+WITH
+    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
+    )
+SELECT DISTINCT
+    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'
+    END AS comparison
+FROM t;
 ```
 
 <!-- tabs:end -->

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

@@ -90,8 +90,30 @@ With he same formula for the average salary comparison in February, the result i
 
 ### **SQL**
 
-```
-
+```sql
+# Write your MySQL query statement below
+WITH
+    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
+    )
+SELECT DISTINCT
+    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'
+    END AS comparison
+FROM t;
 ```
 
 <!-- tabs:end -->

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

@@ -0,0 +1,23 @@
+# Write your MySQL query statement below
+WITH
+    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
+    )
+SELECT DISTINCT
+    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'
+    END AS comparison
+FROM t;