LeetCode #855. Exam Room

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本文介绍了一个算法，用于解决考场中学生自动寻找最远距离座位的问题。通过使用集合数据结构，算法能够高效地找到并返回最优座位，同时在学生离开时更新座位状态。

题目描述：

In an exam room, there are N seats in a single row, numbered 0, 1, 2, ..., 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. (Also, if no one is in the room, then the student sits at seat number 0.)

Return a class ExamRoom(int N) that exposes two functions: ExamRoom.seat() returning an int representing what seat the student sat in, and ExamRoom.leave(int p) representing that the student in seat number p now leaves the room. It is guaranteed that any calls to ExamRoom.leave(p) have a student sitting in seat p.

Example 1:

Input: ["ExamRoom","seat","seat","seat","seat","leave","seat"], [[10],[],[],[],[],[4],[]]
Output: [null,0,9,4,2,null,5]
Explanation:
ExamRoom(10) -> null
seat() -> 0, no one is in the room, then the student sits at seat number 0.
seat() -> 9, the student sits at the last seat number 9.
seat() -> 4, the student sits at the last seat number 4.
seat() -> 2, the student sits at the last seat number 2.
leave(4) -> null
seat() -> 5, the student sits at the last seat number 5.
Note:

1 <= N <= 10^9
ExamRoom.seat() and ExamRoom.leave() will be called at most 10^4 times across all test cases.
Calls to ExamRoom.leave(p) are guaranteed to have a student currently sitting in seat number p.

class ExamRoom {
public:
    ExamRoom(int N) {
        n=N;
    }
    
    int seat() {
        if(s.empty())
        {
            s.insert(0);
            return 0;
        }
        bool first=s.count(0);
        bool last=s.count(n-1);
        int pre=0, result=0, max_dist=0;
        // 遍历set时，需要单独考虑两个端点，当它们没有占用时可以考虑
        // 而且计算两个端点到其他点的距离时不需要除二
        for(const int& i:s)
        {
            int dist=(pre==0&&!first)?i-pre:(i-pre)/2;
            if(dist>max_dist)
            {
                max_dist=dist;
                if(pre==0&&!first) result=0;
                else result=(i+pre)/2;
            }
            pre=i;
        }
        if(n-1-pre>max_dist&&!last) result=n-1;
        s.insert(result);
        return result;
    }
    
    void leave(int p) {
        s.erase(p);
    }

private:
    int n;
    set<int> s;
};