feat: update sql solutions to lc problems

solution/0500-0599/0577.Employee Bonus/README.md

@@ -52,13 +52,12 @@ empId 是这张表单的主关键字
 ### **SQL**
 
 ```sql
-SELECT
-    e.name,
-    b.bonus
+# Write your MySQL query statement below
+SELECT name, bonus
 FROM
-    Employee AS e
-    LEFT JOIN Bonus AS b ON e.empid = b.empid
-WHERE b.bonus < 1000 OR b.bonus IS NULL;
+    Employee
+    LEFT JOIN Bonus USING (empId)
+WHERE ifnull(bonus, 0) < 1000;
 ```
 
 <!-- tabs:end -->

solution/0500-0599/0577.Employee Bonus/README_EN.md

@@ -81,13 +81,12 @@ Bonus table:
 ### **SQL**
 
 ```sql
-SELECT
-    e.name,
-    b.bonus
+# Write your MySQL query statement below
+SELECT name, bonus
 FROM
-    Employee AS e
-    LEFT JOIN Bonus AS b ON e.empid = b.empid
-WHERE b.bonus < 1000 OR b.bonus IS NULL;
+    Employee
+    LEFT JOIN Bonus USING (empId)
+WHERE ifnull(bonus, 0) < 1000;
 ```
 
 <!-- tabs:end -->

solution/0500-0599/0577.Employee Bonus/Solution.sql

@@ -1,7 +1,6 @@
-SELECT
-    e.name,
-    b.bonus
+# Write your MySQL query statement below
+SELECT name, bonus
 FROM
-    Employee AS e
-    LEFT JOIN Bonus AS b ON e.empid = b.empid
-WHERE b.bonus < 1000 OR b.bonus IS NULL;
+    Employee
+    LEFT JOIN Bonus USING (empId)
+WHERE ifnull(bonus, 0) < 1000;

solution/0500-0599/0578.Get Highest Answer Rate Question/README.md

@@ -73,11 +73,28 @@ SurveyLog table:
 ### **SQL**
 
 ```sql
+# Write your MySQL query statement below
 SELECT question_id AS survey_log
-FROM   SurveyLog
-GROUP  BY 1
-ORDER  BY SUM(action = 'answer') / SUM(action = 'show') DESC
-LIMIT  1;
+FROM SurveyLog
+GROUP BY 1
+ORDER BY sum(action = 'answer') / sum(action = 'show') DESC, 1
+LIMIT 1;
+```
+
+```sql
+WITH
+    T AS (
+        SELECT
+            question_id AS survey_log,
+            (sum(action = 'answer') OVER (PARTITION BY question_id)) / (
+                sum(action = 'show') OVER (PARTITION BY question_id)
+            ) AS ratio
+        FROM SurveyLog
+    )
+SELECT survey_log
+FROM T
+ORDER BY ratio DESC, 1
+LIMIT 1;
 ```
 
 <!-- tabs:end -->

solution/0500-0599/0578.Get Highest Answer Rate Question/README_EN.md

@@ -64,11 +64,28 @@ Question 285 has the highest answer rate.</pre>
 ### **SQL**
 
 ```sql
+# Write your MySQL query statement below
 SELECT question_id AS survey_log
-FROM   SurveyLog
-GROUP  BY 1
-ORDER  BY SUM(action = 'answer') / SUM(action = 'show') DESC
-LIMIT  1;
+FROM SurveyLog
+GROUP BY 1
+ORDER BY sum(action = 'answer') / sum(action = 'show') DESC, 1
+LIMIT 1;
+```
+
+```sql
+WITH
+    T AS (
+        SELECT
+            question_id AS survey_log,
+            (sum(action = 'answer') OVER (PARTITION BY question_id)) / (
+                sum(action = 'show') OVER (PARTITION BY question_id)
+            ) AS ratio
+        FROM SurveyLog
+    )
+SELECT survey_log
+FROM T
+ORDER BY ratio DESC, 1
+LIMIT 1;
 ```
 
 <!-- tabs:end -->

solution/0500-0599/0578.Get Highest Answer Rate Question/Solution.sql

@@ -1,5 +1,13 @@
-SELECT question_id AS survey_log
-FROM SurveyLog
-GROUP BY 1
-ORDER BY SUM(action = 'answer') / SUM(action = 'show') DESC
+WITH
+    T AS (
+        SELECT
+            question_id AS survey_log,
+            (sum(action = 'answer') OVER (PARTITION BY question_id)) / (
+                sum(action = 'show') OVER (PARTITION BY question_id)
+            ) AS ratio
+        FROM SurveyLog
+    )
+SELECT survey_log
+FROM T
+ORDER BY ratio DESC, 1
 LIMIT 1;

solution/0500-0599/0579.Find Cumulative Salary of an Employee/README.md

@@ -109,4 +109,28 @@ WHERE
 ORDER BY id, month DESC;
 ```
 
+```sql
+# Write your MySQL query statement below
+WITH
+    T AS (
+        SELECT
+            id,
+            month,
+            sum(salary) OVER (
+                PARTITION BY id
+                ORDER BY month
+                RANGE 2 PRECEDING
+            ) AS salary,
+            rank() OVER (
+                PARTITION BY id
+                ORDER BY month DESC
+            ) AS rk
+        FROM Employee
+    )
+SELECT id, month, salary
+FROM T
+WHERE rk > 1
+ORDER BY 1, 2 DESC;
+```
+
 <!-- tabs:end -->

solution/0500-0599/0579.Find Cumulative Salary of an Employee/README_EN.md

@@ -134,4 +134,28 @@ WHERE
 ORDER BY id, month DESC;
 ```
 
+```sql
+# Write your MySQL query statement below
+WITH
+    T AS (
+        SELECT
+            id,
+            month,
+            sum(salary) OVER (
+                PARTITION BY id
+                ORDER BY month
+                RANGE 2 PRECEDING
+            ) AS salary,
+            rank() OVER (
+                PARTITION BY id
+                ORDER BY month DESC
+            ) AS rk
+        FROM Employee
+    )
+SELECT id, month, salary
+FROM T
+WHERE rk > 1
+ORDER BY 1, 2 DESC;
+```
+
 <!-- tabs:end -->

solution/0500-0599/0579.Find Cumulative Salary of an Employee/Solution.sql

@@ -1,19 +1,21 @@
 # Write your MySQL query statement below
-SELECT
-    id,
-    month,
-    sum(salary) OVER (
-        PARTITION BY id
-        ORDER BY month
-        RANGE 2 PRECEDING
-    ) AS Salary
-FROM employee
-WHERE
-    (id, month) NOT IN (
+WITH
+    T AS (
         SELECT
             id,
-            max(month)
+            month,
+            sum(salary) OVER (
+                PARTITION BY id
+                ORDER BY month
+                RANGE 2 PRECEDING
+            ) AS salary,
+            rank() OVER (
+                PARTITION BY id
+                ORDER BY month DESC
+            ) AS rk
         FROM Employee
-        GROUP BY id
     )
-ORDER BY id, month DESC;
+SELECT id, month, salary
+FROM T
+WHERE rk > 1
+ORDER BY 1, 2 DESC;

solution/0500-0599/0580.Count Student Number in Departments/README.md

@@ -84,14 +84,18 @@ Department 表:
 ### **SQL**
 
 ```sql
-SELECT
-    department.dept_name,
-    COUNT(student.dept_id) AS student_number
+# Write your MySQL query statement below
+WITH
+    S AS (
+        SELECT dept_id, count(1) AS cnt
+        FROM Student
+        GROUP BY dept_id
+    )
+SELECT dept_name, ifnull(cnt, 0) AS student_number
 FROM
-    Student
-    RIGHT JOIN Department ON student.dept_id = department.dept_id
-GROUP BY dept_name
-ORDER BY student_number DESC, dept_name;
+    S
+    RIGHT JOIN Department USING (dept_id)
+ORDER BY 2 DESC, 1;
 ```
 
 <!-- tabs:end -->

solution/0500-0599/0580.Count Student Number in Departments/README_EN.md

@@ -81,14 +81,18 @@ Department table:
 ### **SQL**
 
 ```sql
-SELECT
-    department.dept_name,
-    COUNT(student.dept_id) AS student_number
+# Write your MySQL query statement below
+WITH
+    S AS (
+        SELECT dept_id, count(1) AS cnt
+        FROM Student
+        GROUP BY dept_id
+    )
+SELECT dept_name, ifnull(cnt, 0) AS student_number
 FROM
-    Student
-    RIGHT JOIN Department ON student.dept_id = department.dept_id
-GROUP BY dept_name
-ORDER BY student_number DESC, dept_name;
+    S
+    RIGHT JOIN Department USING (dept_id)
+ORDER BY 2 DESC, 1;
 ```
 
 <!-- tabs:end -->

solution/0500-0599/0580.Count Student Number in Departments/Solution.sql

@@ -1,8 +1,12 @@
-SELECT
-    department.dept_name,
-    COUNT(student.dept_id) AS student_number
+# Write your MySQL query statement below
+WITH
+    S AS (
+        SELECT dept_id, count(1) AS cnt
+        FROM Student
+        GROUP BY dept_id
+    )
+SELECT dept_name, ifnull(cnt, 0) AS student_number
 FROM
-    Student
-    RIGHT JOIN Department ON student.dept_id = department.dept_id
-GROUP BY dept_name
-ORDER BY student_number DESC, dept_name;
+    S
+    RIGHT JOIN Department USING (dept_id)
+ORDER BY 2 DESC, 1;

solution/0500-0599/0584.Find Customer Referee/README.md

@@ -45,19 +45,10 @@
 ### **SQL**
 
 ```sql
-SELECT
-    name
+# Write your MySQL query statement below
+SELECT name
 FROM Customer
-WHERE referee_id != 2 OR referee_id IS NULL;
-```
-
-MySQL 可使用 `IFNULL()`：
-
-```sql
-SELECT
-    name
-FROM customer
-WHERE IFNULL(referee_id, 0) != 2;
+WHERE ifnull(referee_id, 0) != 2;
 ```
 
 <!-- tabs:end -->

solution/0500-0599/0584.Find Customer Referee/README_EN.md

@@ -60,19 +60,10 @@ Customer table:
 ### **SQL**
 
 ```sql
-SELECT
-    name
+# Write your MySQL query statement below
+SELECT name
 FROM Customer
-WHERE referee_id != 2 OR referee_id IS NULL;
-```
-
-MySQL can use `IFNULL()`:
-
-```sql
-SELECT
-    name
-FROM customer
-WHERE IFNULL(referee_id, 0) != 2;
+WHERE ifnull(referee_id, 0) != 2;
 ```
 
 <!-- tabs:end -->

solution/0500-0599/0584.Find Customer Referee/Solution.sql

@@ -1,4 +1,4 @@
-SELECT
-    name
+# Write your MySQL query statement below
+SELECT name
 FROM Customer
-WHERE referee_id != 2 OR referee_id IS NULL;
+WHERE ifnull(referee_id, 0) != 2;