1602. Round Robin Desk
Design a help desk that serves customers in rounds. RoundRobinDesk() starts empty and keeps customer ids in a circular order with a pointer to the customer who is served next.
boolean add(int id)returns false ifidis already present. Otherwise it placesiddirectly before the pointer in the circle (so it is the last to be reached in the current round), returns true, and if the desk was empty the pointer now sits on it.boolean remove(int id)returns false ifidis absent; otherwise it unlinks it and returns true. If it was under the pointer, the pointer moves to the next customer.int serveNext()returns the id under the pointer and moves the pointer forward one step, or -1 when empty.
Every operation must run in O(1) using a circular doubly linked list plus a hash map.
Example 1
- Input:
- operations = ["RoundRobinDesk","add","add","add","serveNext","serveNext","add","serveNext","serveNext","serveNext"]arguments = [[],[5],[7],[9],[],[],[3],[],[],[]]
- Output:
- [null,true,true,true,5,7,true,9,5,7]
- Explanation:
The ring is 5, 7, 9; serving gives 5 then 7, and the new id 3 lands just before the pointer at 9, so 9, 5, 7 are served next.
Example 2
- Input:
- operations = ["RoundRobinDesk","serveNext","remove","add","add","serveNext","serveNext","remove","remove","serveNext"]arguments = [[],[],[1],[4],[4],[],[],[4],[4],[]]
- Output:
- [null,-1,false,true,false,4,4,true,false,-1]
- Explanation:
Operations on an empty desk give -1 and false; the duplicate add of 4 returns false.
Constraints
0 ≤ id ≤ 109; at most 2000 calls in total.
How this problem is judged
- Answers
- Your answer must match exactly. Numbers compare by value, so 2 and 2.0 are equal.
- Input
- Each case is an operation log.
operationsnames the class first and then each method call;argumentsholds the arguments for each, in the same order. Your answer is one list with a result per operation -nullfor the constructor and for methods that return nothing.
Expected complexity
- Time
- O(1) per call
- Space
- O(n)