{ "data": { "question": { "questionId": "732", "questionFrontendId": "732", "categoryTitle": "Algorithms", "boundTopicId": 1829, "title": "My Calendar III", "titleSlug": "my-calendar-iii", "content": "
A k
-booking happens when k
events have some non-empty intersection (i.e., there is some time that is common to all k
events.)
You are given some events [startTime, endTime)
, after each given event, return an integer k
representing the maximum k
-booking between all the previous events.
Implement the MyCalendarThree
class:
MyCalendarThree()
Initializes the object.int book(int startTime, int endTime)
Returns an integer k
representing the largest integer such that there exists a k
-booking in the calendar.\n
Example 1:
\n\n\nInput\n["MyCalendarThree", "book", "book", "book", "book", "book", "book"]\n[[], [10, 20], [50, 60], [10, 40], [5, 15], [5, 10], [25, 55]]\nOutput\n[null, 1, 1, 2, 3, 3, 3]\n\nExplanation\nMyCalendarThree myCalendarThree = new MyCalendarThree();\nmyCalendarThree.book(10, 20); // return 1\nmyCalendarThree.book(50, 60); // return 1\nmyCalendarThree.book(10, 40); // return 2\nmyCalendarThree.book(5, 15); // return 3\nmyCalendarThree.book(5, 10); // return 3\nmyCalendarThree.book(25, 55); // return 3\n\n\n\n
\n
Constraints:
\n\n0 <= startTime < endTime <= 109
400
calls will be made to book
.当 k
个日程安排有一些时间上的交叉时(例如 k
个日程安排都在同一时间内),就会产生 k
次预订。
给你一些日程安排 [start, end)
,请你在每个日程安排添加后,返回一个整数 k
,表示所有先前日程安排会产生的最大 k
次预订。
实现一个 MyCalendarThree
类来存放你的日程安排,你可以一直添加新的日程安排。
MyCalendarThree()
初始化对象。int book(int start, int end)
返回一个整数 k
,表示日历中存在的 k
次预订的最大值。\n\n
示例:
\n\n\n输入:\n[\"MyCalendarThree\", \"book\", \"book\", \"book\", \"book\", \"book\", \"book\"]\n[[], [10, 20], [50, 60], [10, 40], [5, 15], [5, 10], [25, 55]]\n输出:\n[null, 1, 1, 2, 3, 3, 3]\n\n解释:\nMyCalendarThree myCalendarThree = new MyCalendarThree();\nmyCalendarThree.book(10, 20); // 返回 1 ,第一个日程安排可以预订并且不存在相交,所以最大 k 次预订是 1 次预订。\nmyCalendarThree.book(50, 60); // 返回 1 ,第二个日程安排可以预订并且不存在相交,所以最大 k 次预订是 1 次预订。\nmyCalendarThree.book(10, 40); // 返回 2 ,第三个日程安排 [10, 40) 与第一个日程安排相交,所以最大 k 次预订是 2 次预订。\nmyCalendarThree.book(5, 15); // 返回 3 ,剩下的日程安排的最大 k 次预订是 3 次预订。\nmyCalendarThree.book(5, 10); // 返回 3\nmyCalendarThree.book(25, 55); // 返回 3\n\n\n
\n\n
提示:
\n\n0 <= start < end <= 109
book
函数最多不超过 400
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