脉脉maimai面试死亡日记

刚刚结束，真的死得透透的。

前面八股啥的还要我下去了解分层，了解他们的意义。

先说后面俩SQL题吧

求互相关注的对数

我读题不仔细，没看到被关注者是数组！结果傻眼了，面试官问我爆炸函数，我只能说练习中没用到！

sql 复制代码

我感觉这样就可以了，但是HIVE一直报错explode，我输了。。。
WITH exploded_fans AS (
  SELECT from_user, explode(to_user_array) AS to_user FROM fans
)
SELECT COUNT(*) AS mutual_follows_count
FROM exploded_fans ef1
JOIN exploded_fans ef2
  ON ef1.from_user = ef2.to_user
  AND ef1.to_user = ef2.from_user;

连续登录

这个只做出来了row number，而且还是没写对，partition by一开始没写上！后面提醒我连续的事，傻眼了。彻底傻眼。

sql 复制代码

应该可以吧
WITH ranked_logs AS (
    SELECT 
        uid,
        dt,
        ROW_NUMBER() OVER (PARTITION BY uid ORDER BY dt) AS row_num
    FROM user_log
),
date_diff AS (
    SELECT 
        a.uid,
        a.dt,
        DATEDIFF(a.dt, b.dt) AS diff
    FROM ranked_logs a
    JOIN ranked_logs b
        ON a.uid = b.uid
        AND a.row_num = b.row_num + 1
)
SELECT DISTINCT uid
FROM date_diff
where diff = 1
GROUP BY uid
HAVING COUNT(*) = 6;

sql 复制代码

有一个用户粉丝表fans，表中有两个字段：
from_user 关注者 用户ID，
to_user_array 被关注者的一个数组，关注者的粉丝ID集。
如果两个人互相关注，则为一对，求互相关注的对数。


1 2,3
2 1
3 1

# split
# exp
# t t1 join t t2 on t1 in t2.to_user_array


现有一张用户日志表 user_log ，表中有2个字段：
uid 用户id
dt 用户登陆日期
求连续登录7天的用户
# CET
# 连续 !
with t as (
    select
    uid,
    dt,
    row number() over(partition by uid order by dt asc) as rk
from user_log
)
select uid from t where rk >= 7