LeetCode（数据库）- 寻找没有被执行的任务对

-- 解决方案(1)
with recursive t(task_id, subtask_id) as (
    SELECT task_id, subtasks_count FROM Tasks
    UNION ALL
    SELECT task_id, subtask_id-1 FROM t where subtask_id-1>0
)
SELECT * FROM t left join Executed using(task_id, subtask_id) 
WHERE Executed.subtask_id is null 
ORDER BY task_id, subtask_id
-- 解决方案(2)
WITH RECURSIVE t AS(
    SELECT 1 num
    UNION ALL
    SELECT num + 1 FROM t WHERE num < (SELECT MAX(subtasks_count) maxn FROM Tasks)
)
SELECT task_id, num subtask_id
FROM Tasks JOIN t
WHERE subtasks_count >= num AND (task_id, num) NOT IN (SELECT * FROM Executed)
ORDER BY task_id, num