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4 | 4 | importjava.util.LinkedList; |
5 | 5 | importjava.util.Map; |
6 | 6 |
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7 | | -/** |
8 | | - * 1396. Design Underground System |
9 | | - * |
10 | | - * Implement the class UndergroundSystem that supports three methods: |
11 | | - * |
12 | | - * 1. checkIn(int id, string stationName, int t) |
13 | | - * A customer with id card equal to id, gets in the station stationName at time t. |
14 | | - * A customer can only be checked into one place at a time. |
15 | | - * 2. checkOut(int id, string stationName, int t) |
16 | | - * A customer with id card equal to id, gets out from the station stationName at time t. |
17 | | - * 3. getAverageTime(string startStation, string endStation) |
18 | | - * Returns the average time to travel between the startStation and the endStation. |
19 | | - * The average time is computed from all the previous traveling from startStation to endStation that happened directly. |
20 | | - * Call to getAverageTime is always valid. |
21 | | - * 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. |
22 | | - * |
23 | | - * Example 1: |
24 | | - * Input |
25 | | - * ["UndergroundSystem","checkIn","checkIn","checkIn","checkOut","checkOut","checkOut","getAverageTime","getAverageTime","checkIn","getAverageTime","checkOut","getAverageTime"] |
26 | | - * [[],[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"]] |
27 | | - * Output |
28 | | - * [null,null,null,null,null,null,null,14.0,11.0,null,11.0,null,12.0] |
29 | | - * |
30 | | - * Explanation |
31 | | - * UndergroundSystem undergroundSystem = new UndergroundSystem(); |
32 | | - * undergroundSystem.checkIn(45, "Leyton", 3); |
33 | | - * undergroundSystem.checkIn(32, "Paradise", 8); |
34 | | - * undergroundSystem.checkIn(27, "Leyton", 10); |
35 | | - * undergroundSystem.checkOut(45, "Waterloo", 15); |
36 | | - * undergroundSystem.checkOut(27, "Waterloo", 20); |
37 | | - * undergroundSystem.checkOut(32, "Cambridge", 22); |
38 | | - * undergroundSystem.getAverageTime("Paradise", "Cambridge"); // return 14.0. There was only one travel from "Paradise" (at time 8) to "Cambridge" (at time 22) |
39 | | - * undergroundSystem.getAverageTime("Leyton", "Waterloo"); // return 11.0. 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 = 11.0 |
40 | | - * undergroundSystem.checkIn(10, "Leyton", 24); |
41 | | - * undergroundSystem.getAverageTime("Leyton", "Waterloo"); // return 11.0 |
42 | | - * undergroundSystem.checkOut(10, "Waterloo", 38); |
43 | | - * undergroundSystem.getAverageTime("Leyton", "Waterloo"); // return 12.0 |
44 | | - * |
45 | | - * Constraints: |
46 | | - * There will be at most 20000 operations. |
47 | | - * 1 <= id, t <= 10^6 |
48 | | - * All strings consist of uppercase, lowercase English letters and digits. |
49 | | - * 1 <= stationName.length <= 10 |
50 | | - * Answers within 10^-5 of the actual value will be accepted as correct. |
51 | | - * */ |
52 | 7 | publicclass_1396 { |
53 | 8 | publicstaticclassSolution1 { |
54 | 9 | publicclassUndergroundSystem { |
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