文章目錄

• • 1. 題目
  • 2. 解題

1. 題目

Numbers 表保存數字的值及其頻率。

+----------+-------------+ | Number | Frequency | +----------+-------------| | 0 | 7 | | 1 | 1 | | 2 | 3 | | 3 | 1 | +----------+-------------+

在此表中，數字為 0, 0, 0, 0, 0, 0, 0, 1, 2, 2, 2, 3，所以中位數是 (0 + 0) / 2 = 0。

+--------+ | median | +--------| | 0.0000 | +--------+

請編寫一個查詢來查找所有數字的中位數并將結果命名為 median 。

來源：力扣（LeetCode）
鏈接：https://leetcode-cn.com/problems/find-median-given-frequency-of-numbers
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2. 解題

• 求出正序、逆序的前綴頻數和

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]]}

• 選出正反序中間位置的數（前綴頻數大于等于一半的第一個滿足的數），最后求平均

# Write your MySQL query statement below select avg(Number) median from ((select Numberfrom(select Number, sum(Frequency) over(order by Number asc) presumfrom Numbers) t1where presum >= ceil((select sum(Frequency) from Numbers)/2)order by Numberlimit 1)union all(select Numberfrom(select Number, sum(Frequency) over(order by Number desc) presumfrom Numbers) t2where presum >= ceil((select sum(Frequency) from Numbers)/2)order by Number desclimit 1) ) tmp

評論區簡潔解法

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 () cntfrom Numbers ) t where c1 >= cnt/2 and c2 >= cnt/2

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