1. 题目
Numbers 表保存数字的值及其频率。
+----------+-------------+
| Number | Frequency |
+----------+-------------|
| 0 | 7 |
| 1 | 1 |
| 2 | 3 |
| 3 | 1 |
+----------+-------------+
在此表中，数字为 0, 0, 0, 0, 0, 0, 0, 1, 2, 2, 2, 3，所以中位数是 (0 + 0) / 2 = 0。
+--------+
| median |
+--------|
| 0.0000 |
+--------+
请编写一个查询来查找所有数字的中位数并将结果命名为 median 。
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/find-median-given-frequency-of-numbers
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2. 解题
- 求出正序、逆序的前缀频数和
select Number, sum(Frequency) over(order by Number asc) presum
from Numbers
{"headers": ["Number", "presum"],
"values": [[0, 7], [1, 8], [2, 11], [3, 12]]}
select Number, sum(Frequency) over(order by Number desc) presum
from Numbers
{"headers": ["Number", "presum"],
"values": [[3, 1], [2, 4], [1, 5], [0, 12]]}
- 选出正反序中间位置的数（前缀频数大于等于一半的第一个满足的数），最后求平均
# Write your MySQL query statement below
select avg(Number) median
from
(
(
select Number
from
(
select Number, sum(Frequency) over(order by Number asc) presum
from Numbers
) t1
where presum >= ceil((select sum(Frequency) from Numbers)/2)
order by Number
limit 1
)
union all
(
select Number
from
(
select Number, sum(Frequency) over(order by Number desc) presum
from Numbers
) t2
where presum >= ceil((select sum(Frequency) from Numbers)/2)
order by Number desc
limit 1
)
) tmp
评论区简洁解法
select avg(Number) median
from
(
select Number,
sum(Frequency) over (order by Number asc) c1,
sum(Frequency) over (order by Number desc) c2,
sum(Frequency) over () cnt
from Numbers
) t
where c1 >= cnt/2 and c2 >= cnt/2
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