動機

就很直接，沒有什麼魔法

Problem

There is an exam room with n seats in a single row labeled from 0 to n - 1.

When a student enters the room, they must sit in the seat that maximizes the distance to the closest person. If there are multiple such seats, they sit in the seat with the lowest number. If no one is in the room, then the student sits at seat number 0.

Design a class that simulates the mentioned exam room.

Implement the ExamRoom class:

  • ExamRoom(int n) Initializes the object of the exam room with the number of the seats n.
  • int seat() Returns the label of the seat at which the next student will set.
  • void leave(int p) Indicates that the student sitting at seat p will leave the room. It is guaranteed that there will be a student sitting at seat p.

 

Example 1:

Input["ExamRoom", "seat", "seat", "seat", "seat", "leave", "seat"][[10], [], [], [], [], [4], []]Output[Algorithm, Leetcode, 0, 9, 4, 2, null, 5]ExplanationExamRoom examRoom = new ExamRoom(10);examRoom.seat(); // return 0, no one is in the room, then the student sits at seat number 0.examRoom.seat(); // return 9, the student sits at the last seat number 9.examRoom.seat(); // return 4, the student sits at the last seat number 4.examRoom.seat(); // return 2, the student sits at the last seat number 2.examRoom.leave(4);examRoom.seat(); // return 5, the student sits at the last seat number 5.

 

Constraints:

  • 1 <= n <= 109
  • It is guaranteed that there is a student sitting at seat p.
  • At most 104 calls will be made to seat and leave.

Sol

每次就是看要插入的點與前一個點的距離，找最大的

但要注意，最後有一個N的位置要看

from sortedcontainers import SortedList
class ExamRoom:
    
    def __init__(self, n: int):
        self.l = SortedList()
        self.n = n
    def seat(self) -> int:
        if len(self.l) == 0:
            ret = 0
        else:
            d = self.l[0] # 到前一個位置的距離
            ret = 0
            for a,b in zip(self.l, self.l[1:]):
                if (b-a)//2 > d:
                    ret = (a+b)//2
                    d = (b-a)//2
            if self.n-1-self.l[-1] > d:
                ret = self.n-1
        self.l.add(ret)
        return ret

    def leave(self, p: int) -> None:
        self.l.remove(p)