35 - 最后一个能进入巴士的人

-- sum(weight) over(order by turn) as total,根据turn升序，再求前面数的和
select
    person_name 
from
    (
        select
            person_name,sum(weight) over(order by turn) as total
        from
            Queue
    ) new_Queue
where
    total<=1000
order by 
    total desc
limit 1;