https://leetcode.cn/problems/students-with-invalid-departments/
这个就不能用大表left join小表了，大表left join小表旨在求出那些在大表中，都能在小表中找到对应的匹配项的记录。这题是反过来，要求找不到匹配项，即 is null的记录，所以得小表left join大表。
SQL
# Write your MySQL query statement below
SELECT A.id,A.name
FROM Students AS A
LEFT JOIN Departments AS B
ON A.department_id = B.id
WHERE B.name is NULL