题目地址

https://leetcode.cn/problems/consecutive-transactions-with-increasing-amounts/

代码

# Write your MySQL query statement below

with t1 as(
    select 
    *

    # --------------------------只需要改动这里的逻辑，其他不要动。注意里面的语句是“顺序执行的”----------------------

    -- 如果customer_id切换了
    ,(case when (@prev_customer_id is null or @prev_customer_id<>customer_id) then @interval_id := @interval_id +1 end) -- 这句看你需要，如果跨顾客需要interval_id额外间隔1，你应该保留这一句。这句注释或者不注释都能AC
    ,(case when @prev_customer_id is null or @prev_customer_id<>customer_id  then @reg:=0 end)
    ,(case when @prev_customer_id is null or @prev_customer_id<>customer_id  then @prev_amoutn:=0 end)
    ,(case when @prev_customer_id is null or @prev_customer_id<>customer_id  then @consecutive_start:=transaction_date end)
    ,(case when @prev_customer_id is null or @prev_customer_id<>customer_id  then @consecutive_end:=transaction_date end)
    ,(case when @prev_customer_id is null or @prev_customer_id<>customer_id  then @prev_transaction_date:= transaction_date - interval 1 day end)  -- 当前日的上一日，哨兵
    
    , @prev_transaction_date as prev_transaction_date  -- 把这个用户变量@prev_transaction_date存到列里。



     -- 这是维护一个【严格递增连续子序列的长度】
     ,(case when amount>@prev_amount and @prev_transaction_date + interval 1 day =transaction_date then @reg:=@reg+1 else @reg:=1 end)

     ,@reg as reg   -- 把这个用户变量@reg存到列里。

     -- 更新一下consecutive_start
     ,(case when customer_id=@prev_customer_id and amount>@prev_amount and @prev_transaction_date + interval 1 day =transaction_date 
        then @consecutive_start:=@consecutive_start 
     else @consecutive_start:=transaction_date end) as consecutive_start


     -- 不管什么情况，这么更新consecutive_end总是对的
     ,@consecutive_end:=transaction_date as consecutive_end


     -- 更新一下interval_id
     ,(case when customer_id=@prev_customer_id and  amount>@prev_amount and @prev_transaction_date + interval 1 day =transaction_date 
        then @interval_id:=@interval_id 
     else @interval_id:=@interval_id+1 end) as interval_id



    -- 准备好处理下一行了，一些@prev_*用户变量更新一下
    ,@prev_amount:=amount
    ,@prev_customer_id:=customer_id
    ,@prev_transaction_date:=transaction_date
    ,@prev_amoutn:=amount

    # --------------------------只需要改动这里的逻辑，其他不要动。注意里面的语句是“顺序执行的”----------------------

    from 
    (select * from Transactions order by customer_id asc,  transaction_date asc) sorted_ticks,
    (select @reg:=0 -- 某用户每日购买量严格增加的天数（连续日）
    ,@prev_amount:=null, @prev_customer_id:=null, @prev_transaction_date:=null,@prev_amoutn:=null
    ,@consecutive_start:=null , @consecutive_end:=null,@interval_id := 0)vars
)


-- 对于每一个interval_id，找到reg最大的那一行，就是该区间最后的统计信息

select customer_id,consecutive_start,consecutive_end
from (select *, row_number() over(partition by interval_id order by reg desc) as rn from t1) tmp1
where rn=1 and reg>=3
order by customer_id,consecutive_start,consecutive_end



-- 下面注释掉的代码是用来调试的
# select 
# customer_id,transaction_date,reg,amount,prev_transaction_date,interval_id
# ,consecutive_start,consecutive_end
# from t1
# order by customer_id asc, transaction_date asc