Implement a MyCalendarThree class to store your events. A new event can alwaysbe 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.

Your class will be called like this: 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 most 400.
• In calls to MyCalendarThree.book(start, end), start and end are integers in the range [0, 10^9].

分析

与之前的 My Calendar 和  My Calendar II不同，这道题要求给出当前Calendar 中最大的重复次数。这道题如果不知道使用map的巧妙解法，写起来会相当费劲。使用map<int, int>来记录当前的时间段，只不过每次插入[start, end)时，map[start] ++, map[end] --，之后从小打到遍历key，计算value的和，value的和的最大值就代表了重复时间段的最多次数。

举例说明：[5, 15), [10, 20), [10, 18)

{5, 1}, {15, -1}

{5, 1}, {10, 1}, {15, -1}, {20, -1}

{5, 1}, {10, 2}, {15, -1}, {18, -1}, {20, -1}

那么计算出value的和最大值为3。

Code

class MyCalendarThree {
public:
    MyCalendarThree() {
        
    }
    
    int book(int start, int end) {
        if (m.find(start) == m.end())
            m[start] = 0;
        if (m.find(end) == m.end())
            m[end] = 0;
        
        m[start] ++;
        m[end] --;
        map<int, int>::iterator iter;
        int count = 0;
        int res = 0;
        for (iter = m.begin(); iter!= m.end(); iter ++)
        {
            count += iter->second;
            res = max(res, count);
        }
        return res;
    }
    
    map<int, int> m;
};

运行效率

Runtime: 144 ms, faster than 33.22% of C++ online submissions for My Calendar III.

Memory Usage: 25.1 MB, less than 72.05% of C++ online submissions for My Calendar III.