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855. 考场就座

在考场里,一排有 N 个座位,分别编号为 0, 1, 2, ..., N-1 。

当学生进入考场后,他必须坐在能够使他与离他最近的人之间的距离达到最大化的座位上。如果有多个这样的座位,他会坐在编号最小的座位上。(另外,如果考场里没有人,那么学生就坐在 0 号座位上。)

返回 ExamRoom(int N) 类,它有两个公开的函数:其中,函数 ExamRoom.seat() 会返回一个 int (整型数据),代表学生坐的位置;函数 ExamRoom.leave(int p) 代表坐在座位 p 上的学生现在离开了考场。每次调用 ExamRoom.leave(p) 时都保证有学生坐在座位 p 上。

 

示例:

输入:["ExamRoom","seat","seat","seat","seat","leave","seat"], [[10],[],[],[],[],[4],[]]
输出:[null,0,9,4,2,null,5]
解释:
ExamRoom(10) -> null
seat() -> 0,没有人在考场里,那么学生坐在 0 号座位上。
seat() -> 9,学生最后坐在 9 号座位上。
seat() -> 4,学生最后坐在 4 号座位上。
seat() -> 2,学生最后坐在 2 号座位上。
leave(4) -> null
seat() -> 5,学生最后坐在 5 号座位上。

 

提示:

  1. 1 <= N <= 10^9
  2. 在所有的测试样例中 ExamRoom.seat() 和 ExamRoom.leave() 最多被调用 10^4 次。
  3. 保证在调用 ExamRoom.leave(p) 时有学生正坐在座位 p 上。

相似题目

到最近的人的最大距离

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上次编辑到这里,代码来自缓存 点击恢复默认模板
class ExamRoom { public: ExamRoom(int n) { } int seat() { } void leave(int p) { } }; /** * Your ExamRoom object will be instantiated and called as such: * ExamRoom* obj = new ExamRoom(n); * int param_1 = obj->seat(); * obj->leave(p); */

python3 解法, 执行用时: 228 ms, 内存消耗: 21.3 MB, 提交时间: 2022-12-30 10:06:12

from sortedcontainers import SortedList


class ExamRoom:

    def __init__(self, n: int):
        def dist(x):
            l, r = x
            return r - l - 1 if l == -1 or r == n else (r - l) >> 1

        self.n = n
        self.ts = SortedList(key=lambda x: (-dist(x), x[0]))
        self.left = {}
        self.right = {}
        self.add((-1, n))

    def seat(self) -> int:
        s = self.ts[0]
        p = (s[0] + s[1]) >> 1
        if s[0] == -1:
            p = 0
        elif s[1] == self.n:
            p = self.n - 1
        self.delete(s)
        self.add((s[0], p))
        self.add((p, s[1]))
        return p

    def leave(self, p: int) -> None:
        l, r = self.left[p], self.right[p]
        self.delete((l, p))
        self.delete((p, r))
        self.add((l, r))

    def add(self, s):
        self.ts.add(s)
        self.left[s[1]] = s[0]
        self.right[s[0]] = s[1]

    def delete(self, s):
        self.ts.remove(s)
        self.left.pop(s[1])
        self.right.pop(s[0])


# Your ExamRoom object will be instantiated and called as such:
# obj = ExamRoom(n)
# param_1 = obj.seat()
# obj.leave(p)

golang 解法, 执行用时: 32 ms, 内存消耗: 8 MB, 提交时间: 2022-12-30 10:05:42

type ExamRoom struct {
	rbt   *redblacktree.Tree
	left  map[int]int
	right map[int]int
	n     int
}

func Constructor(n int) ExamRoom {
	dist := func(s []int) int {
		if s[0] == -1 || s[1] == n {
			return s[1] - s[0] - 1
		}
		return (s[1] - s[0]) >> 1
	}
	cmp := func(a, b interface{}) int {
		x, y := a.([]int), b.([]int)
		d1, d2 := dist(x), dist(y)
		if d1 == d2 {
			return x[0] - y[0]
		}
		return d2 - d1
	}
	this := ExamRoom{redblacktree.NewWith(cmp), map[int]int{}, map[int]int{}, n}
	this.add([]int{-1, n})
	return this
}

func (this *ExamRoom) Seat() int {
	s := this.rbt.Left().Key.([]int)
	p := (s[0] + s[1]) >> 1
	if s[0] == -1 {
		p = 0
	} else if s[1] == this.n {
		p = this.n - 1
	}
	this.del(s)
	this.add([]int{s[0], p})
	this.add([]int{p, s[1]})
	return p
}

func (this *ExamRoom) Leave(p int) {
	l, _ := this.left[p]
	r, _ := this.right[p]
	this.del([]int{l, p})
	this.del([]int{p, r})
	this.add([]int{l, r})
}

func (this *ExamRoom) add(s []int) {
	this.rbt.Put(s, struct{}{})
	this.left[s[1]] = s[0]
	this.right[s[0]] = s[1]
}

func (this *ExamRoom) del(s []int) {
	this.rbt.Remove(s)
	delete(this.left, s[1])
	delete(this.right, s[0])
}

/**
 * Your ExamRoom object will be instantiated and called as such:
 * obj := Constructor(n);
 * param_1 := obj.Seat();
 * obj.Leave(p);
 */

java 解法, 执行用时: 23 ms, 内存消耗: 46.4 MB, 提交时间: 2022-12-30 10:04:54

class ExamRoom {
    private TreeSet<int[]> ts = new TreeSet<>((a, b) -> {
        int d1 = dist(a), d2 = dist(b);
        return d1 == d2 ? a[0] - b[0] : d2 - d1;
    });
    private Map<Integer, Integer> left = new HashMap<>();
    private Map<Integer, Integer> right = new HashMap<>();
    private int n;

    public ExamRoom(int n) {
        this.n = n;
        add(new int[] {-1, n});
    }

    public int seat() {
        int[] s = ts.first();
        int p = (s[0] + s[1]) >> 1;
        if (s[0] == -1) {
            p = 0;
        } else if (s[1] == n) {
            p = n - 1;
        }
        del(s);
        add(new int[] {s[0], p});
        add(new int[] {p, s[1]});
        return p;
    }

    public void leave(int p) {
        int l = left.get(p), r = right.get(p);
        del(new int[] {l, p});
        del(new int[] {p, r});
        add(new int[] {l, r});
    }

    private int dist(int[] s) {
        int l = s[0], r = s[1];
        return l == -1 || r == n ? r - l - 1 : (r - l) >> 1;
    }

    private void add(int[] s) {
        ts.add(s);
        left.put(s[1], s[0]);
        right.put(s[0], s[1]);
    }

    private void del(int[] s) {
        ts.remove(s);
        left.remove(s[1]);
        right.remove(s[0]);
    }
}

/**
 * Your ExamRoom object will be instantiated and called as such:
 * ExamRoom obj = new ExamRoom(n);
 * int param_1 = obj.seat();
 * obj.leave(p);
 */

java 解法, 执行用时: 603 ms, 内存消耗: 43.4 MB, 提交时间: 2022-12-30 10:03:23

class ExamRoom {
    int N;
    TreeSet<Integer> students;

    public ExamRoom(int N) {
        this.N = N;
        students = new TreeSet();
    }

    public int seat() {
        //Let's determine student, the position of the next
        //student to sit down.
        int student = 0;
        if (students.size() > 0) {
            //Tenatively, dist is the distance to the closest student,
            //which is achieved by sitting in the position 'student'.
            //We start by considering the left-most seat.
            int dist = students.first();
            Integer prev = null;
            for (Integer s: students) {
                if (prev != null) {
                    //For each pair of adjacent students in positions (prev, s),
                    //d is the distance to the closest student;
                    //achieved at position prev + d.
                    int d = (s - prev) / 2;
                    if (d > dist) {
                        dist = d;
                        student = prev + d;
                    }
                }
                prev = s;
            }

            //Considering the right-most seat.
            if (N - 1 - students.last() > dist)
                student = N - 1;
        }

        //Add the student to our sorted TreeSet of positions.
        students.add(student);
        return student;
    }

    public void leave(int p) {
        students.remove(p);
    }
}

/**
 * Your ExamRoom object will be instantiated and called as such:
 * ExamRoom obj = new ExamRoom(n);
 * int param_1 = obj.seat();
 * obj.leave(p);
 */

python3 解法, 执行用时: 6228 ms, 内存消耗: 18.1 MB, 提交时间: 2022-12-30 10:03:00

class ExamRoom(object):
    def __init__(self, N):
        self.N = N
        self.students = []

    def seat(self):
        # Let's determine student, the position of the next
        # student to sit down.
        if not self.students:
            student = 0
        else:
            # Tenatively, dist is the distance to the closest student,
            # which is achieved by sitting in the position 'student'.
            # We start by considering the left-most seat.
            dist, student = self.students[0], 0
            for i, s in enumerate(self.students):
                if i:
                    prev = self.students[i-1]
                    # For each pair of adjacent students in positions (prev, s),
                    # d is the distance to the closest student;
                    # achieved at position prev + d.
                    d = (s - prev) // 2
                    if d > dist:
                        dist, student = d, prev + d

            # Considering the right-most seat.
            d = self.N - 1 - self.students[-1]
            if d > dist:
                student = self.N - 1

        # Add the student to our sorted list of positions.
        bisect.insort(self.students, student)
        return student

    def leave(self, p):
        self.students.remove(p)

# Your ExamRoom object will be instantiated and called as such:
# obj = ExamRoom(n)
# param_1 = obj.seat()
# obj.leave(p)

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