LeetCode（数据库）- 寻找面试候选人

Posted 2021-12-25 程序员牧码

题目链接：点击打开链接

题目大意：略。

解题思路：略。

AC 代码

-- 解决方案(1)
# db 临时表：列传行
with db as (
select *
from (
    select gold_medal as id,contest_id from contests
    union all
    select silver_medal as id,contest_id from contests
    union all
    select bronze_medal as id,contest_id from contests) as t
order by id, contest_id) # 这里需要排序，不然变量遍历会出错


select name, mail
from(
    #条件二：三场及更多不同的比赛中赢得金牌
    select gold_medal as user_id
    from contests
    group by gold_medal
    having count(distinct contest_id) >= 3
    
    union
    
    #条件一：该用户在连续三场及更多比赛中赢得奖牌
    select distinct id as user_id
    from(
        select 
            id,
            if(id = @lid, if(contest_id = @lcontest_id + 1, @cnt:=@cnt, @cnt:=@cnt+1), @cnt:=1) as cnt,
            @lid := id,
            @lcontest_id := contest_id
        from db, (select @lid:=-999, @lcontest_id:=-999, @cnt:=1) as u) as p
    group by id,cnt
    having count(cnt) >= 3) as y

left join users m on y.user_id = m.user_id

-- 解决方案(2)
WITH t AS(SELECT gold_medal user_id, contest_id FROM Contests
UNION ALL
SELECT silver_medal, contest_id FROM Contests
UNION ALL
SELECT bronze_medal, contest_id FROM Contests),

tt AS(SELECT t_1.user_id
FROM t t_1 JOIN t t_2 ON t_1.user_id = t_2.user_id AND t_1.contest_id + 1 = t_2.contest_id 
JOIN t t_3 ON t_2.user_id = t_3.user_id AND t_2.contest_id + 1 = t_3.contest_id
UNION
SELECT gold_medal user_id FROM Contests GROUP BY gold_medal HAVING COUNT(*) > 2)

SELECT name, mail
FROM tt JOIN Users USING(user_id)