How to Design Underground System using Several Hash Maps?
- 时间：2020-09-10 13:03:17
- 分类：网络文摘
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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 12.00000Example 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 6.66667Constraints:
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.Hints:
Use two hash tables. The first to save the check-in time for a customer and the second to update the total time between two stations.
Design a Underground System using C++ unordered_map
We can use hashmaps to remember a number of things: time in/out, place in/out, the number of occurences between same stops, and the latest average. In C++, we use unordered_map to store the key-value pairs (internally implemented as a hash map).
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 | class UndergroundSystem { public: UndergroundSystem() { } void checkIn(int id, string stationName, int t) { timeIn[id] = t; placeIn[id] = stationName; } void checkOut(int id, string stationName, int t) { timeOut[id] = t; placeOut[id] = stationName; string startStation = placeIn[id]; int c = count[placeIn[id]][stationName] ++; average[startStation][stationName] = (average[startStation][stationName] * c + timeOut[id] - timeIn[id]) / (c + 1); } double getAverageTime(string startStation, string endStation) { return average[startStation][endStation]; } private: unordered_map<int, int> timeIn; unordered_map<int, int> timeOut; unordered_map<int, string> placeIn; unordered_map<int, string> placeOut; unordered_map<string, unordered_map<string, int>> count; unordered_map<string, unordered_map<string, double>> average; }; |
class UndergroundSystem {
public:
UndergroundSystem() {
}
void checkIn(int id, string stationName, int t) {
timeIn[id] = t;
placeIn[id] = stationName;
}
void checkOut(int id, string stationName, int t) {
timeOut[id] = t;
placeOut[id] = stationName;
string startStation = placeIn[id];
int c = count[placeIn[id]][stationName] ++;
average[startStation][stationName] = (average[startStation][stationName] * c + timeOut[id] - timeIn[id]) / (c + 1);
}
double getAverageTime(string startStation, string endStation) {
return average[startStation][endStation];
}
private:
unordered_map<int, int> timeIn;
unordered_map<int, int> timeOut;
unordered_map<int, string> placeIn;
unordered_map<int, string> placeOut;
unordered_map<string, unordered_map<string, int>> count;
unordered_map<string, unordered_map<string, double>> average;
};The getAverageTime, checkIn and checkOut will be all O(1) constant. The average can be computed based on the following:
1 2 | average = (average * count + latest) / (count + 1); count ++; |
average = (average * count + latest) / (count + 1); count ++;
The space complexity (requirement) is of course O(N) linear to the number of the check-in/check-outs.
–EOF (The Ultimate Computing & Technology Blog) —
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