首页 > 其他分享> > [LeetCode] 1396. Design Underground System

[LeetCode] 1396. Design Underground System

2020-10-07 05:31:24 作者：互联网

Implement the class UndergroundSystem that supports three methods:

1. checkIn(int id, string stationName, int t)

A customer with id card equal to id, gets in the station stationName at time t.
A customer can only be checked into one place at a time.

2. checkOut(int id, string stationName, int t)

A customer with id card equal to id, gets out from the station stationName at time t.

3. getAverageTime(string startStation, string endStation)

Returns the average time to travel between the startStation and the endStation.
The average time is computed from all the previous traveling from startStation to endStation that happened directly.
Call to getAverageTime is always valid.

You can assume all calls to checkIn and checkOut methods are consistent. That is, if a customer gets in at time t1 at some station, then it gets out at time t2 with t2 > t1. All events happen in chronological order.

Example 1:

Input
["UndergroundSystem","checkIn","checkIn","checkIn","checkOut","checkOut","checkOut","getAverageTime","getAverageTime","checkIn","getAverageTime","checkOut","getAverageTime"]
[[],[45,"Leyton",3],[32,"Paradise",8],[27,"Leyton",10],[45,"Waterloo",15],[27,"Waterloo",20],[32,"Cambridge",22],["Paradise","Cambridge"],["Leyton","Waterloo"],[10,"Leyton",24],["Leyton","Waterloo"],[10,"Waterloo",38],["Leyton","Waterloo"]]

Output
[null,null,null,null,null,null,null,14.00000,11.00000,null,11.00000,null,12.00000]

Explanation
UndergroundSystem undergroundSystem = new UndergroundSystem();
undergroundSystem.checkIn(45, "Leyton", 3);
undergroundSystem.checkIn(32, "Paradise", 8);
undergroundSystem.checkIn(27, "Leyton", 10);
undergroundSystem.checkOut(45, "Waterloo", 15);
undergroundSystem.checkOut(27, "Waterloo", 20);
undergroundSystem.checkOut(32, "Cambridge", 22);
undergroundSystem.getAverageTime("Paradise", "Cambridge");       // return 14.00000. There was only one travel from "Paradise" (at time 8) to "Cambridge" (at time 22)
undergroundSystem.getAverageTime("Leyton", "Waterloo");          // return 11.00000. There were two travels from "Leyton" to "Waterloo", a customer with id=45 from time=3 to time=15 and a customer with id=27 from time=10 to time=20. So the average time is ( (15-3) + (20-10) ) / 2 = 11.00000
undergroundSystem.checkIn(10, "Leyton", 24);
undergroundSystem.getAverageTime("Leyton", "Waterloo");          // return 11.00000
undergroundSystem.checkOut(10, "Waterloo", 38);
undergroundSystem.getAverageTime("Leyton", "Waterloo");          // return 12.00000

Example 2:

Input
["UndergroundSystem","checkIn","checkOut","getAverageTime","checkIn","checkOut","getAverageTime","checkIn","checkOut","getAverageTime"]
[[],[10,"Leyton",3],[10,"Paradise",8],["Leyton","Paradise"],[5,"Leyton",10],[5,"Paradise",16],["Leyton","Paradise"],[2,"Leyton",21],[2,"Paradise",30],["Leyton","Paradise"]]

Output
[null,null,null,5.00000,null,null,5.50000,null,null,6.66667]

Explanation
UndergroundSystem undergroundSystem = new UndergroundSystem();
undergroundSystem.checkIn(10, "Leyton", 3);
undergroundSystem.checkOut(10, "Paradise", 8);
undergroundSystem.getAverageTime("Leyton", "Paradise"); // return 5.00000
undergroundSystem.checkIn(5, "Leyton", 10);
undergroundSystem.checkOut(5, "Paradise", 16);
undergroundSystem.getAverageTime("Leyton", "Paradise"); // return 5.50000
undergroundSystem.checkIn(2, "Leyton", 21);
undergroundSystem.checkOut(2, "Paradise", 30);
undergroundSystem.getAverageTime("Leyton", "Paradise"); // return 6.66667

Constraints:

There will be at most 20000 operations.
1 <= id, t <= 10^6
All strings consist of uppercase, lowercase English letters and digits.
1 <= stationName.length <= 10
Answers within 10^-5 of the actual value will be accepted as correct.

设计地铁系统。

请你实现一个类 UndergroundSystem ，它支持以下 3 种方法：

1. checkIn(int id, string stationName, int t)

编号为 id 的乘客在 t 时刻进入地铁站 stationName 。
一个乘客在同一时间只能在一个地铁站进入或者离开。
2. checkOut(int id, string stationName, int t)

编号为 id 的乘客在 t 时刻离开地铁站 stationName 。
3. getAverageTime(string startStation, string endStation)

返回从地铁站 startStation 到地铁站 endStation 的平均花费时间。
平均时间计算的行程包括当前为止所有从 startStation 直接到达 endStation 的行程。
调用 getAverageTime 时，询问的路线至少包含一趟行程。
你可以假设所有对 checkIn 和 checkOut 的调用都是符合逻辑的。也就是说，如果一个顾客在 t1 时刻到达某个地铁站，那么他离开的时间 t2 一定满足 t2 > t1 。所有的事件都按时间顺序给出。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/design-underground-system
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

这道题我们需要两个hashmap，一个叫做checkinMap，记录每个乘客checkin的细节，一个叫做checkoutMap，记录每个route的细节，route包含了上车地点和下车地点。其中checkinMap是记录乘客的id，<开始车站，时间>；checkoutMap是记录route，<时间，count>。这里我们需要用到Java 8的一个新功能叫做Pair<，他可以存一些键值对，用法跟hashmap几乎一样。

checkIn()函数。当有乘客check in，我们就把他的ID，<上车地点stationName，时间戳t>记录在checkinMap里。

checkOut()函数。当有乘客下车，我们首先去checkinMap拿到他的上车信息，这样才能拼接他的路径。路径由上车地点 + "_" + 下车地点组成。下车地点由checkOut()函数提供了。同时因为乘客下车了，我们需要结算他的乘车时间 = 下车时间 - 上车时间。

得到路径和乘车时间之后，我们需要把这两个信息放入checkoutMap里。因为有可能有多人有相同的路径，存住这个信息，才能算接下来的getAverageTime。

getAverageTime()函数。将给的上车地点和下车地点拼接成route，把这个route当做key去checkoutMap里找是否存在，若存在，则拿出那个对应的pair，pair里存的是所有当前route的花费时间的总和，除以次数，即得到平均时间。

时间 - N/A

空间 - N/A

Java实现

 1 class UndergroundSystem {
 2     // 乘客的id，<开始车站，时间>
 3     HashMap<Integer, Pair<String, Integer>> checkinMap = new HashMap<>();
 4     // route，<总计时间，相同route的出现次数>
 5     HashMap<String, Pair<Integer, Integer>> checkoutMap = new HashMap<>();
 6 
 7     public UndergroundSystem() {
 8 
 9     }
10 
11     public void checkIn(int id, String stationName, int t) {
12         // 乘客的id，<开始车站，时间>
13         checkinMap.put(id, new Pair<>(stationName, t));
14     }
15 
16     public void checkOut(int id, String stationName, int t) {
17         // 拿到乘客的<开始车站，时间>
18         Pair<String, Integer> checkInDetail = checkinMap.get(id);
19         // 拼接route
20         String route = checkInDetail.getKey() + "_" + stationName;
21         // 花费时间
22         int totalTime = t - checkInDetail.getValue();
23         Pair<Integer, Integer> checkout = checkoutMap.getOrDefault(route, new Pair<>(0, 0));
24         checkoutMap.put(route, new Pair<>(checkout.getKey() + totalTime, checkout.getValue() + 1));
25     }
26 
27     public double getAverageTime(String startStation, String endStation) {
28         String route = startStation + "_" + endStation;
29         Pair<Integer, Integer> checkout = checkoutMap.get(route);
30         return (double) checkout.getKey() / checkout.getValue();
31     }
32 }
33 
34 /**
35  * Your UndergroundSystem object will be instantiated and called as such:
36  * UndergroundSystem obj = new UndergroundSystem();
37  * obj.checkIn(id,stationName,t);
38  * obj.checkOut(id,stationName,t);
39  * double param_3 = obj.getAverageTime(startStation,endStation);
40  */

LeetCode 题目总结

标签：checkIn,Leyton,getAverageTime,Design,undergroundSystem,Underground,1396,checkOut
来源： https://www.cnblogs.com/cnoodle/p/13776187.html