My Calendar II

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最新推荐文章于 2020-08-10 11:12:17 发布

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本文详细解析了LeetCode上MyCalendarTwo问题的解决方案，通过使用set和hashtable数据结构，有效地处理了事件预订中可能出现的三重预订冲突，确保了在不超过两次预订重叠的情况下，新事件能够成功添加。

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731. My Calendar II

Implement a MyCalendarTwo class to store your events. A new event can be added if adding the event will not cause a triple booking.

Your class will have one method, book(int start, int end). Formally, this represents a booking on the half open interval [start, end), the range of real numbers x such that start <= x < end.

A triple booking happens when three events have some non-empty intersection (ie., there is some time that is common to all 3 events.)

For each call to the method MyCalendar.book, return true if the event can be added to the calendar successfully without causing a triple booking. Otherwise, return false and do not add the event to the calendar.

Your class will be called like this: MyCalendar cal = new MyCalendar();MyCalendar.book(start, end)

Example 1:

MyCalendar();
MyCalendar.book(10, 20); // returns true
MyCalendar.book(50, 60); // returns true
MyCalendar.book(10, 40); // returns true
MyCalendar.book(5, 15); // returns false
MyCalendar.book(5, 10); // returns true
MyCalendar.book(25, 55); // returns true
Explanation: 
The first two events can be booked.  The third event can be double booked.
The fourth event (5, 15) can't be booked, because it would result in a triple booking.
The fifth event (5, 10) can be booked, as it does not use time 10 which is already double booked.
The sixth event (25, 55) can be booked, as the time in [25, 40) will be double booked with the third event;
the time [40, 50) will be single booked, and the time [50, 55) will be double booked with the second event.

Note:

The number of calls to MyCalendar.book per test case will be at most 1000.
In calls to MyCalendar.book(start, end), start and end are integers in the range [0, 10^9].

1.解析

题目大意，判断间隔是否能插入(不存在3处重叠的区域)。

2.分析

这道题是My Calendar I的进化版，其实就是在原来的基础上将重复的区域改成可以至多允许存在两处重复的区域，所以，关键之处在于如何判断待插入的区间出现3处重复区域。我们可以用hashtable将之前出现过的重复区域保存下来，当插入一个新的间隔时，判断是否在之前已经出现过的重复区域，若出现，则说明会出现3处重复区域。思路参考@Grandyang

class MyCalendarTwo {
public:
    MyCalendarTwo() {
        
    }
    
    bool book(int start, int end) {
        for (auto& cur: overlap){
            if (end <= cur.first || cur.second <= start) continue;
            return false; //存在3个triple时间
        }
        for (auto& cur: interval){
            if (end <= cur.first || cur.second <= start) continue;
            overlap.insert({max(cur.first, start), min(cur.second, end)}); //增加重叠区域
        }
        interval.insert({start, end});
        return true;
    }
    set<pair<int, int>> interval, overlap;
};

/**
 * Your MyCalendarTwo object will be instantiated and called as such:
 * MyCalendarTwo* obj = new MyCalendarTwo();
 * bool param_1 = obj->book(start,end);
 */