[LeetCode] 1396. Design Underground System
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 stationstationName
at timet
. - 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 stationstationName
at timet
.
3. getAverageTime(string startStation, string endStation)
- Returns the average time to travel between the
startStation
and theendStation
. - The average time is computed from all the previous traveling from
startStation
toendStation
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
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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 */