SQL面试题练习 —— 取出累计值与1000差值最小的记录

题目来源：滴滴

1 题目

已知有表 t_cost_detail 包含 id 和 money 两列，id 为自增，请累加计算money 值，并求出累加值与 1000 差值最小的记录。

+-----+--------+
| id  | money  |
+-----+--------+
| 1   | 200    |
| 2   | 300    |
| 3   | 200    |
| 4   | 100    |
| 5   | 150    |
| 6   | 80     |
| 7   | 100    |
| 8   | 200    |
+-----+--------+

样例结果

+-----+--------+
| id  | money  |
+-----+--------+
| 6   | 80     |
+-----+--------+

2 建表语句

CREATE TABLE IF NOT EXISTS t_cost_detail (
    id bigint, --id
    money bigint -- money
)
ROW FORMAT DELIMITED
FIELDS TERMINATED BY ','
STORED AS ORC;

insert into t_cost_detail(id, money) values
(1,200),
(2,300),
(3,200),
(4,100),
(5,150),
(6,80),
(7,100),
(8,200);

3 题解

select
    id,
    money
from
    (
    select
        id,
        money,
        row_number()over(order by abs_diff) as rn
    from
        (
        select
            id,
            money,
            sum(money) over(order by id) as sum_money,
            abs(sum(money) over(order by id)-1000) as abs_diff
        from t_cost_detail
        ) t
    ) tt
where rn = 1;