feat: add solutions to lc problems: No.0614,1270

* No.0614.Second Degree Follower
* No.1270.All People Report to the Given Manager

Author: yanglbme

solution/0600-0699/0614.Second Degree Follower/README.md

@@ -56,7 +56,14 @@
 ### **SQL**
 
 ```sql
-
+# Write your MySQL query statement below
+SELECT  f1.follower
+       ,COUNT(DISTINCT f2.follower) AS num
+FROM follow f1
+JOIN follow f2
+ON f1.follower = f2.followee
+GROUP BY  f1.follower
+ORDER BY f1.follower
 ```
 
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solution/0600-0699/0614.Second Degree Follower/README_EN.md

@@ -67,7 +67,14 @@ User Alice has 1 follower. Alice is not a second-degree follower because she doe
 ### **SQL**
 
 ```sql
-
+# Write your MySQL query statement below
+SELECT  f1.follower
+       ,COUNT(DISTINCT f2.follower) AS num
+FROM follow f1
+JOIN follow f2
+ON f1.follower = f2.followee
+GROUP BY  f1.follower
+ORDER BY f1.follower
 ```
 
 <!-- tabs:end -->

solution/0600-0699/0614.Second Degree Follower/Solution.sql

@@ -0,0 +1,8 @@
+# Write your MySQL query statement below
+SELECT  f1.follower
+       ,COUNT(DISTINCT f2.follower) AS num
+FROM follow f1
+JOIN follow f2
+ON f1.follower = f2.followee
+GROUP BY  f1.follower
+ORDER BY f1.follower

solution/1200-1299/1270.All People Report to the Given Manager/README.md

@@ -73,15 +73,14 @@ employee_id 是 3, 8 ，9 的职员不会直接或间接的汇报给公司 CEO
 
 ```sql
 # Write your MySQL query statement below
-
-SELECT e1.employee_id
-FROM   employees e1
-JOIN   employees e2
-JOIN   employees e3
-ON     e1.manager_id=e2.employee_id
-AND    e2.manager_id=e3.employee_id
-where  e3.manager_id=1
-AND    e1.employee_id!=1;
+SELECT  e1.employee_id
+FROM Employees e1
+JOIN Employees e2
+ON e1.manager_id = e2.employee_id
+JOIN Employees e3
+ON e2.manager_id = e3.employee_id
+WHERE e1.employee_id != 1
+AND e3.manager_id = 1
 ```
 
 <!-- tabs:end -->

solution/1200-1299/1270.All People Report to the Given Manager/README_EN.md

@@ -73,14 +73,15 @@ The employees with employee_id 3, 8, and 9 do not report their work to the head
 ```sql
 # Write your MySQL query statement below
 
-SELECT e1.employee_id
-FROM   employees e1
-JOIN   employees e2
-JOIN   employees e3
-ON     e1.manager_id=e2.employee_id
-AND    e2.manager_id=e3.employee_id
-where  e3.manager_id=1
-AND    e1.employee_id!=1;
+# Write your MySQL query statement below
+SELECT  e1.employee_id
+FROM Employees e1
+JOIN Employees e2
+ON e1.manager_id = e2.employee_id
+JOIN Employees e3
+ON e2.manager_id = e3.employee_id
+WHERE e1.employee_id != 1
+AND e3.manager_id = 1
 ```
 
 <!-- tabs:end -->

solution/1200-1299/1270.All People Report to the Given Manager/Solution.sql

@@ -1,9 +1,9 @@
 # Write your MySQL query statement below
-SELECT e1.employee_id
-FROM   employees e1
-JOIN   employees e2
-JOIN   employees e3
-ON     e1.manager_id=e2.employee_id
-AND    e2.manager_id=e3.employee_id
-where  e3.manager_id=1
-AND    e1.employee_id!=1;
+SELECT  e1.employee_id
+FROM Employees e1
+JOIN Employees e2
+ON e1.manager_id = e2.employee_id
+JOIN Employees e3
+ON e2.manager_id = e3.employee_id
+WHERE e1.employee_id != 1
+AND e3.manager_id = 1