Second Highest Salary --leetCode

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 query should return 200 as the second highest salary. If there is no second highest salary, then the query should return null.

+---------------------+
| SecondHighestSalary |
+---------------------+
| 200                 |
+---------------------+

https://leetcode.com/problems/second-highest-salary/description/

solution:

本想找第二高的就是子查询找到 两个最高的 而后取最小的那个
select Salary as SecondHighestSalary from
(select * from Employee order by Salary desc limit 2) temp
order by Salary  asc limit 1
可是执行 告诉我结果是错的  原来题目的要求中没有第二高的就返回 null

看来转换思路很重要呀，当时想按照上面的方式 怎么处理子查询只有一条的状况呀 真是陷入了本身的坑
而后按照取到max最大的，从比最大的小中的最大的，若是我要取第三大的 是否是比较坑
SELECT MAX(Salary) as SecondHighestSalary
FROM Employee
WHERE Salary < (SELECT MAX(Salary) FROM Employee);


而后去看看网上的大神都是怎么处理的 这边UNION ALL 和 UNION是同样的吧
SELECT Salary as SecondHighestSalary FROM Employee GROUP BY Salary
UNION ALL (SELECT NULL AS SecondHighestSalary)
ORDER BY SecondHighestSalary DESC LIMIT 1,1;
若是是我 mysql估计就会卡在这边怎么处理null的问题出不来了
SELECT Salary as SecondHighestSalary FROM Employee GROUP BY Salary  DESC LIMIT 1,1

下面这个max的用法 和上面的殊途同归
SELECT MAX(Salary) as SecondHighestSalary  FROM Employee
WHERE Salary NOT IN
(SELECT MAX(Salary) FROM Employee);

后面这我的写的真直接呀
SELECT MAX(Salary) as SecondHighestSalary FROM Employee E1
WHERE 1 =
(SELECT COUNT(DISTINCT(E2.Salary)) FROM Employee E2
WHERE E2.Salary > E1.Salary);

git地址：https://github.com/woshiyexinjie/leetcode-xin