update: 7 solutions

Author: hijiangtao

README.md

@@ -2,12 +2,14 @@
 
 This is the solution collection of my LeetCode problems, most of them are programmed in JavaScript.
 
-Progress: 4/
+Progress: 7/
 
 | ID | Title | Solution | Difficulty |
 |---| ----- | -------- | ---------- |
 |1|[Two Sum](https://leetcode.com/problems/two-sum/) | [JavaScript](./src/two-sum/res.js)|Easy|
 |2|[Add Two Numbers](https://leetcode.com/problems/add-two-numbers/) | [JavaScript](./src/add-two-numbers/res.js)|Medium|
 |175|[Combine Two Tables](https://leetcode.com/problems/combine-two-tables/)| [SQL](./src/combine-two-tables/res.txt)|Easy|
+|176|[Second Highest Salary](https://leetcode.com/problems/second-highest-salary/)| [SQL](./src/second-highest-salary/res.txt)|Easy|
 |181|[Employees Earning More Than Their Managers](https://leetcode.com/problems/employees-earning-more-than-their-managers/) | [SQL](./src/employees-earning-more-than-their-managers/res.txt)|Easy|
+|182|[Duplicate Emails](https://leetcode.com/problems/duplicate-emails/) | [SQL](./src/duplicate-emails/res.txt)|Easy|
 |434|[Number of Segments in a String](https://leetcode.com/problems/number-of-segments-in-a-string/) | [JavaScript](./src/number-of-segments-in-a-string/res.js)|Easy|

src/duplicate-emails/res.txt

@@ -0,0 +1,22 @@
+-- Write a SQL query to find all duplicate emails in a table named Person.
+
+-- +----+---------+
+-- | Id | Email   |
+-- +----+---------+
+-- | 1  | [email protected] |
+-- | 2  | [email protected] |
+-- | 3  | [email protected] |
+-- +----+---------+
+-- For example, your query should return the following for the above table:
+
+-- +---------+
+-- | Email   |
+-- +---------+
+-- | [email protected] |
+-- +---------+
+-- Note: All emails are in lowercase.
+
+# Write your MySQL query statement below
+SELECT t.eml AS 'Email' FROM 
+(SELECT count(Email) AS 'cnt', Email AS 'eml' FROM Person GROUP BY Email) as t 
+WHERE t.cnt>1;

src/second-highest-salary/res.txt

@@ -0,0 +1,23 @@
+-- Write a SQL query to get the second highest salary from the Employee table.
+
+-- +----+--------+
+-- | Id | Salary |
+-- +----+--------+
+-- | 1  | 100    |
+-- | 2  | 200    |
+-- | 3  | 300    |
+-- +----+--------+
+-- For example, given the above Employee table, the second highest salary is 200. If there is no second highest salary, then the query should return null.
+
+# Note @hijiangtao
+# You should rename the result header as SecondHighestSalary. 
+# NOT IN condition will consume much more time than < condition
+# Note @hijiangtao
+
+# Write your MySQL query statement below
+SELECT MAX(Salary) AS "SecondHighestSalary" FROM Employee
+WHERE Salary < (SELECT MAX(Salary) FROM Employee )
+
+# Another solution
+SELECT MAX(Salary) AS "SecondHighestSalary" FROM Employee
+WHERE Salary NOT IN (SELECT MAX(Salary) FROM Employee )