Implement a MyCalendarThree
class to store your events. A new event can always be added.
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 K-booking happens when K events have some non-empty intersection (ie., there is some time that is common to all K events.)
For each call to the method MyCalendar.book
, return an integer K
representing the largest integer such that there exists a K
-booking in the calendar.
MyCalendarThree cal = new MyCalendarThree();
MyCalendarThree.book(start, end)
Example 1:
MyCalendarThree(); MyCalendarThree.book(10, 20); // returns 1 MyCalendarThree.book(50, 60); // returns 1 MyCalendarThree.book(10, 40); // returns 2 MyCalendarThree.book(5, 15); // returns 3 MyCalendarThree.book(5, 10); // returns 3 MyCalendarThree.book(25, 55); // returns 3 Explanation: The first two events can be booked and are disjoint, so the maximum K-booking is a 1-booking. The third event [10, 40) intersects the first event, and the maximum K-booking is a 2-booking. The remaining events cause the maximum K-booking to be only a 3-booking. Note that the last event locally causes a 2-booking, but the answer is still 3 because eg. [10, 20), [10, 40), and [5, 15) are still triple booked.
Note:
- The number of calls to
MyCalendarThree.book
per test case will be at most400
. - In calls to
MyCalendarThree.book(start, end)
,start
andend
are integers in the range[0, 10^9]
.
这道题是之前那两道题My Calendar II,My Calendar I的拓展,论坛上有人说这题不应该算是Hard类的,但实际上如果没有之前那两道题做铺垫,直接上这道其实还是还蛮有难度的。这道题博主在做完之前那道,再做这道一下子就做出来了,因为用的就是之前那道My Calendar II的解法二,具体的讲解可以参见那道题,反正博主写完那道题再来做这道题就是秒解啊,参见代码如下:
class MyCalendarThree { public: MyCalendarThree() {} int book(int start, int end) { ++freq[start]; --freq[end]; int cnt = 0, mx = 0; for (auto f : freq) { cnt += f.second; mx = max(mx, cnt); } return mx; } private: map<int, int> freq; };
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