【leetcode】Trips and Users
The Trips table holds all taxi trips. Each trip has a unique Id, while Client_Id and Driver_Id are both foreign keys to the Users_Id at the Users table. Status is an ENUM type of (‘completed’, ‘cancelled_by_driver’, ‘cancelled_by_client’).
+----+-----------+-----------+---------+--------------------+----------+
| Id | Client_Id | Driver_Id | City_Id | Status |Request_at|
+----+-----------+-----------+---------+--------------------+----------+
| 1 | 1 | 10 | 1 | completed |2013-10-01|
| 2 | 2 | 11 | 1 | cancelled_by_driver|2013-10-01|
| 3 | 3 | 12 | 6 | completed |2013-10-01|
| 4 | 4 | 13 | 6 | cancelled_by_client|2013-10-01|
| 5 | 1 | 10 | 1 | completed |2013-10-02|
| 6 | 2 | 11 | 6 | completed |2013-10-02|
| 7 | 3 | 12 | 6 | completed |2013-10-02|
| 8 | 2 | 12 | 12 | completed |2013-10-03|
| 9 | 3 | 10 | 12 | completed |2013-10-03|
| 10 | 4 | 13 | 12 | cancelled_by_driver|2013-10-03|
+----+-----------+-----------+---------+--------------------+----------+
The Users table holds all users. Each user has an unique Users_Id, and Role is an ENUM type of (‘client’, ‘driver’, ‘partner’).
+----------+--------+--------+
| Users_Id | Banned | Role |
+----------+--------+--------+
| 1 | No | client |
| 2 | Yes | client |
| 3 | No | client |
| 4 | No | client |
| 10 | No | driver |
| 11 | No | driver |
| 12 | No | driver |
| 13 | No | driver |
+----------+--------+--------+
Write a SQL query to find the cancellation rate of requests made by unbanned clients between Oct 1, 2013 and Oct 3, 2013. For the above tables, your SQL query should return the following rows with the cancellation rate being rounded to two decimal places.
+------------+-------------------+
| Day | Cancellation Rate |
+------------+-------------------+
| 2013-10-01 | 0.33 |
| 2013-10-02 | 0.00 |
| 2013-10-03 | 0.50 |
+------------+-------------------+ 解题思路:本题属于hard级别,但我个人觉得其难度不大。 题目的意思是要求出2013-10-01到2013-10-03这三天内,每天非Banned用户取消订单的比率,取消订单包括乘客取消和司机取消。 1.取出所有满足条件的记录,这次加了一个按照日期排序,给后续统计使用
select * from Trips a, Users b where a.Client_Id = b.Users_Id and b.Role = 'client' and b.Banned = 'No' and Request_at between '2013-10-01' and '2013-10-03' order by Request_at;
2.依次遍历所有符合条件的记录,引入中间遍历@day(当前记录的日期),@lastday(上一条记录的日期),@snum(完成的记录数),@cnum(被取消的记录数),@totalnum(记录总数)。如果@day和@lastday相等,表示日期在同一天,根据订单状态分别给@snum或@cnum加1,同时给@totalnum加1;
如果不相等,表示日期变换了,如果订单状态成功让@snum = 1,@cnum = 0,如果失败让@snum = 0,@cnum = 1,同时给@totalnum = 1。这也就是为什么在第一步中要按日期排序的原因了,遍历完成后,同一日期的最后一条记录中的@snum,@cnum,@totalnum就是这个日期对应的各状态的数量。
select Request_at,
@lastday:=@day,
case when @day = '' then @day:=Request_at when @day != Request_at then (@day:=Request_at) else @day:=Request_at end,
@snum := if(@lastday = Request_at , if(Status='completed',@snum+1,@snum), if(Status='completed',@snum:=1,@sum:=0)) as success ,
@cnum := if(@lastday = Request_at , if(Status!='completed',@cnum+1,@cnum), if(Status!='completed',@cnum:=1,@csum:=0)) as fail ,
@totalnum := if(@lastday = Request_at , @totalnum+1, @totalnum:=1) as total
from (select Request_at,Status,Client_Id from Trips order by Request_at) a, Users b ,(select @snum:=0,@cnum:=0,@day:='',@lastday:='',@totalnum:=0) c where a.Client_Id = b.Users_Id and b.Role = 'client' and b.Banned = 'No' order by Request_at,total desc;
3.用@cnum除以@totalnum求出商即可
select Request_at as Day,round(fail/total,2) as 'Cancellation Rate' from
(
select Request_at,
@lastday:=@day,
case when @day = '' then @day:=Request_at when @day != Request_at then (@day:=Request_at) else @day:=Request_at end,
@snum := if(@lastday = Request_at , if(Status='completed',@snum+1,@snum), if(Status='completed',@snum:=1,@sum:=0)) as success ,
@cnum := if(@lastday = Request_at , if(Status!='completed',@cnum+1,@cnum), if(Status!='completed',@cnum:=1,@csum:=0)) as fail ,
@totalnum := if(@lastday = Request_at , @totalnum+1, @totalnum:=1) as total
from (select Request_at,Status,Client_Id from Trips order by Request_at) a, Users b ,(select @snum:=0,@cnum:=0,@day:='',@lastday:='',@totalnum:=0) c where a.Client_Id = b.Users_Id and b.Role = 'client' and b.Banned = 'No' and Request_at between '2013-10-01' and '2013-10-03' order by Request_at ,total desc
) d group by Request_at;
【leetcode】Trips and Users的更多相关文章
- 【LeetCode】Minimum Depth of Binary Tree 二叉树的最小深度 java
[LeetCode]Minimum Depth of Binary Tree Given a binary tree, find its minimum depth. The minimum dept ...
- 【Leetcode】Pascal's Triangle II
Given an index k, return the kth row of the Pascal's triangle. For example, given k = 3, Return [1,3 ...
- 53. Maximum Subarray【leetcode】
53. Maximum Subarray[leetcode] Find the contiguous subarray within an array (containing at least one ...
- 27. Remove Element【leetcode】
27. Remove Element[leetcode] Given an array and a value, remove all instances of that value in place ...
- 【刷题】【LeetCode】007-整数反转-easy
[刷题][LeetCode]总 用动画的形式呈现解LeetCode题目的思路 参考链接-空 007-整数反转 方法: 弹出和推入数字 & 溢出前进行检查 思路: 我们可以一次构建反转整数的一位 ...
- 【刷题】【LeetCode】000-十大经典排序算法
[刷题][LeetCode]总 用动画的形式呈现解LeetCode题目的思路 参考链接 000-十大经典排序算法
- 【leetcode】893. Groups of Special-Equivalent Strings
Algorithm [leetcode]893. Groups of Special-Equivalent Strings https://leetcode.com/problems/groups-o ...
- 【leetcode】657. Robot Return to Origin
Algorithm [leetcode]657. Robot Return to Origin https://leetcode.com/problems/robot-return-to-origin ...
- 【leetcode】557. Reverse Words in a String III
Algorithm [leetcode]557. Reverse Words in a String III https://leetcode.com/problems/reverse-words-i ...
随机推荐
- OpenGL_构建GLFW与第一个程序
参考教程:https://learnopengl-cn.github.io/ 这个教程已经给出了很详细的资料,当然我这里是对细节的展示(在Windows上). 首先,你需要准备 VS2017 : ht ...
- (转)C++ bitset用法
今天做题发现要用到bitset,找到一篇介绍的巨好的文章. 转载自:https://www.cnblogs.com/magisk/p/8809922.html C++的 bitset 在 bitset ...
- Linux 常用服务器命令
1.查看端口号是否被占用 netstat -lnp|grep 端口 或 lsof -i :端口 2查看进程对应的端口号 netstat -nap | grep 进程号
- java-selenium定位元素和操作元素
八种定位方式 一.By.id(id):通过ID 属性查找 HTML 源码 <a onclick="return false;" id="lb" name= ...
- Luogu P4902 乘积
题目 我们要求的是 \[ \prod\limits_{i=a}^b\prod\limits_{j=1}^i(\frac ij)^{\lfloor\frac ij\rfloor} \] 先把它拆开 \[ ...
- C++练习 | 单链表的创建与输出(结构体格式)
#include <iostream> #include <stdio.h> using namespace std; #define OK 1 #define ERROR 0 ...
- idea 新建maven项目时,避免每次都需要指定自己的maven目录
01 .File->Other Settings -> Settings for New Project 02. 将Maven home directory目录修改成我们自己安装Maven ...
- 剑指offer-构建乘积数组-数组-python
题目描述 给定一个数组A[0,1,...,n-1],请构建一个数组B[0,1,...,n-1],其中B中的元素B[i]=A[0]*A[1]*...*A[i-1]*A[i+1]*...*A[n-1].不 ...
- php过滤微信昵称中的表情
function filterNickname($nickname) { $nickname = preg_replace('/[\x{1F600}-\x{1F64F}]/u', '', $nickn ...
- 104、验证Swarm数据持久性 (Swarm11)
参考https://www.cnblogs.com/CloudMan6/p/8016994.html 上一节我们成功将 nfs 的volume挂载到 Service上,本节验证 Failover时 ...