This is the solution collection of my LeetCode problems, most of them are programmed in JavaScript.

Progress:7/

Progress:9/

| 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|

|177|[Nth Highest Salary](https://leetcode.com/problems/nth-highest-salary/)|[SQL](./src/nth-highest-salary/res.txt)|Medium|

|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|

|197|[Rising Temperature](https://leetcode.com/problems/rising-temperature/)|[SQL](./src/rising-temperature/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 get the nth highest salary from the Employee table.

-- +----+--------+

-- | Id | Salary |

-- +----+--------+

-- | 1 | 100 |

-- | 2 | 200 |

-- | 3 | 300 |

-- +----+--------+

-- For example, given the above Employee table, the nth highest salary where n = 2 is 200. If there is no nth highest salary, then the query should return null.

CREATE FUNCTION getNthHighestSalary(N INT) RETURNS INT

BEGIN

DECLARE M INT;

SET M=N-1;

RETURN (

# Write your MySQL query statement below.

SELECT DISTINCT Salary FROM Employee ORDER BY Salary DESC LIMIT M,1

);

END

-- Given a Weather table, write a SQL query to find all dates' Ids with higher temperature compared to its previous (yesterday's) dates.

-- +---------+------------+------------------+

-- | Id(INT) | Date(DATE) | Temperature(INT) |

-- +---------+------------+------------------+

-- | 1 | 2015-01-01 | 10 |

-- | 2 | 2015-01-02 | 25 |

-- | 3 | 2015-01-03 | 20 |

-- | 4 | 2015-01-04 | 30 |

-- +---------+------------+------------------+

-- For example, return the following Ids for the above Weather table:

-- +----+

-- | Id |

-- +----+

-- | 2 |

-- | 4 |

-- +----+

# Write your MySQL query statement below

SELECT a.Id AS 'Id' FROM Weather a INNER JOIN Weather b ON a.Temperature > b.Temperature AND DATEDIFF(a.Date, b.Date)=1