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|

-- 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.

# 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;

-- 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 )