[Function add]

1.Add leetcode solutions.

Author: SeanForFun

README.md

@@ -323,6 +323,10 @@
 
 [180. Consecutive Numbers](https://github.com/Seanforfun/Algorithm-and-Leetcode/blob/master/leetcode/180.%20Consecutive%20Numbers.md)
 
+[181. Employees Earning More Than Their Managers](https://github.com/Seanforfun/Algorithm-and-Leetcode/blob/master/leetcode/181.%20EmployeesVEarning%20More%20Than%20Their%20Managers.md)
+
+[182. Duplicate Emails](https://github.com/Seanforfun/Algorithm-and-Leetcode/blob/master/leetcode/182.%20Duplicate%20Emails.md)
+
 ## Algorithm(4th_Edition)
 Reading notes of book Algorithm(4th Algorithm),ISBN: 9787115293800.
 All java realization codes are placed in different packages.

leetcode/181. Employees Earning More Than Their Managers.md

@@ -3,22 +3,26 @@
 ### Question
 The Employee table holds all employees including their managers. Every employee has an Id, and there is also a column for the manager Id.
 
-+----+-------+--------+-----------+
+```
+----+-------+--------+-----------+
 | Id | Name  | Salary | ManagerId |
 +----+-------+--------+-----------+
 | 1  | Joe   | 70000  | 3         |
 | 2  | Henry | 80000  | 4         |
 | 3  | Sam   | 60000  | NULL      |
 | 4  | Max   | 90000  | NULL      |
 +----+-------+--------+-----------+
+```
 
 Given the Employee table, write a SQL query that finds out employees who earn more than their managers. For the above table, Joe is the only employee who earns more than his manager.
 
+```
 +----------+
 | Employee |
 +----------+
 | Joe      |
 +----------+
+```
 
 ### Thinking:
 * Method: 左外连接，速度比where要快。
@@ -53,4 +57,24 @@ where
         where
             a.ManagerId = b.Id
     );
+```
+
+# 二刷
+1. 使用左外连接，这样我们可以合并两张表的内容。
+2. 使用where做出匹配。
+```Sql
+# Write your MySQL query statement below
+SELECT a.Name as Employee
+FROM Employee a
+LEFT JOIN Employee b
+ON a.ManagerId = b.Id
+WHERE a.Salary > b.Salary;
+```
+
+3. 将同一张表使用两次做出匹配。
+```SQL
+# Write your MySQL query statement below
+SELECT a.Name as Employee
+FROM Employee a, Employee b
+where a.ManagerId = b.Id and a.Salary > b.Salary;
 ```

leetcode/182. Duplicate Emails.md

@@ -1,37 +1,25 @@
 ## 181. Employees Earning More Than Their Managers
 
-### Question
-The Employee table holds all employees including their managers. Every employee has an Id, and there is also a column for the manager Id.
-
-+----+-------+--------+-----------+
-| Id | Name  | Salary | ManagerId |
-+----+-------+--------+-----------+
-| 1  | Joe   | 70000  | 3         |
-| 2  | Henry | 80000  | 4         |
-| 3  | Sam   | 60000  | NULL      |
-| 4  | Max   | 90000  | NULL      |
-+----+-------+--------+-----------+
-
-Given the Employee table, a## 182. Duplicate Emails
-
 ### Question
 Write a SQL query to find all duplicate emails in a table named Person.
-
+```
 +----+---------+
 | Id | Email   |
 +----+---------+
 | 1  | a@b.com |
 | 2  | c@d.com |
 | 3  | a@b.com |
 +----+---------+
+```
 
 For example, your query should return the following for the above table:
-
+```
 +---------+
 | Email   |
 +---------+
 | a@b.com |
 +---------+
+```
 
 Note: All emails are in lowercase.
 
@@ -48,45 +36,14 @@ group by
     Email
 having
     count(*) > 1
-```write a SQL query that finds out employees who earn more than their managers. For the above table, Joe is the only employee who earns more than his manager.
-
-+----------+
-| Employee |
-+----------+
-| Joe      |
-+----------+
-
-### Thinking:
-* Method: 左外连接，速度比where要快。
-
-```SQL
-# Write your MySQL query statement below
-select
-    a.Name as Employee
-from Employee a
-left join
-    Employee b
-on
-    a.ManagerID = b.Id
-where
-    a.Salary > b.Salary
 ```
 
-* Method 2:通过查询确定判断条件，这种方法比较慢，经过两次查询，使用了两个session。
-
-```Java
+### 二刷
+1. 使用Email字段作为分组的类别，将count()作为聚合函数。
+```SQL
 # Write your MySQL query statement below
-select
-    name as Employee
-from
-    Employee a
-where
-    a.Salary > (
-        select
-            b.Salary
-        from
-            Employee b
-        where
-            a.ManagerId = b.Id
-    );
+SELECT Email
+FROM Person
+GROUP BY Email
+HAVING  count(*) > 1;
 ```