571. 给定数字的频率查询中位数 难度：困难

1、题目描述

来源：力扣（LeetCode）

2、解题思路

1# 首先，对次数进行累计，计算出数字的总数:
select max(@i:=@i+frequency) as cou from Numbers,(select @i:=0) a

2# 然后，由总数量，算出两个中位数的位置（如果是奇数，则两个位置相等）：
select if (cou%2=0,cou/2+1,cou/2+0.5) as _p1,if (cou%2=0,cou/2,cou/2+0.5) as _p2 from(
select max(@i:=@i+frequency) as cou from Numbers,(select @i:=0) a
)add_count

3# 这一步，联合Numbers表，使用变量，构造出前一个数字的总数累计，和当前数字的总数累计，判断上一步的_p1或_p2是否落在中间，如果是则提取Numbers,否则取0
select if(@m<_p1 and _p1<=@m:=@m+Frequency,Number,0) as m1, if(@n<_p2 and _p2<=@n:=@n+Frequency,Number,0) as m2 from(
select if (cou%2=0,cou/2+1,cou/2+0.5) as _p1,if (cou%2=0,cou/2,cou/2+0.5) as _p2
from(select max(@i:=@i+frequency) as cou
from Numbers,(select @i:=0) a )add_count
)add_p ,(select * from Numbers order by Number) num_order,(select @m:=0,@n:=0)b

因为两个中位数都是0，所以不明显。将变量@m取值也一起查询出来，看的更清楚：
select @m as m_befor,if(@m<_p1 and _p1<=@m:=@m+Frequency,Number,0) as m1,@m as m_after

4# 最后就简单了，因为m1，m2中除了中位数，就是0，所以直接sum
select (sum(m1)+sum(m2))/2 as median

3、提交记录

select (sum(m1)+sum(m2))/2 as median
from(
 	select if(@m<_p1 and _p1<=@m:=@m+Frequency,Number,0) as m1,
 	if(@n<_p2 and _p2<=@n:=@n+Frequency,Number,0) as m2
     from(
  		select if (cou%2=0,cou/2+1,cou/2+0.5) as _p1,if (cou%2=0,cou/2,cou/2+0.5) as _p2
   		from(
			select max(@i:=@i+frequency) as cou
			from Numbers,(select @i:=0) a  
   			)add_count)add_p,
   			(select * from Numbers order by Number) num_order,(select @m:=0,@n:=0)b)add_m

205ms