732 My Calendar III 我的日程安排表 III
Description:
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 [start, end), 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 start, int end) Returns an integer k representing the largest integer such that there exists a k-booking in the calendar.
Example:
Example 1:
Input
["MyCalendarThree", "book", "book", "book", "book", "book", "book"]
[[], [10, 20], [50, 60], [10, 40], [5, 15], [5, 10], [25, 55]]
Output
[null, 1, 1, 2, 3, 3, 3]
Explanation
MyCalendarThree myCalendarThree = new MyCalendarThree();
myCalendarThree.book(10, 20); // return 1, The first event can be booked and is disjoint, so the maximum k-booking is a 1-booking.
myCalendarThree.book(50, 60); // return 1, The second event can be booked and is disjoint, so the maximum k-booking is a 1-booking.
myCalendarThree.book(10, 40); // return 2, The third event [10, 40) intersects the first event, and the maximum k-booking is a 2-booking.
myCalendarThree.book(5, 15); // return 3, The remaining events cause the maximum K-booking to be only a 3-booking.
myCalendarThree.book(5, 10); // return 3
myCalendarThree.book(25, 55); // return 3
Constraints:
0 <= start < end <= 10^9
At most 400 calls will be made to book.
题目描述:
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。
示例 :
输入：
["MyCalendarThree", "book", "book", "book", "book", "book", "book"]
[[], [10, 20], [50, 60], [10, 40], [5, 15], [5, 10], [25, 55]]
输出：
[null, 1, 1, 2, 3, 3, 3]
解释：
MyCalendarThree myCalendarThree = new MyCalendarThree();
myCalendarThree.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。
myCalendarThree.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。
myCalendarThree.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。
myCalendarThree.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。
myCalendarThree.book(5, 10); // 返回 3
myCalendarThree.book(25, 55); // 返回 3
提示:
0 <= start < end <= 10^9
每个测试用例，调用 book 函数最多不超过 400次
思路:
有序字典
参考 LeetCode #731 My Calendar II 我的日程安排表 II
使用 C++ 中的 map 或者 Java 中的 TreeMap 或者 Python 中的 sortedcontainers.SortedDict 实现
记录开始和结束的时间(边界值)
开始时间 + 1, 结束时间 - 1
遍历整个字典累计 count 值
更新最大的 count 值并返回
时间复杂度为 O(n ^ 2), 空间复杂度为 O(n)
代码:
C++:
class MyCalendarThree
{
private:
map<int, int> m;
public:
MyCalendarThree() {}
int book(int start, int end)
{
++m[start];
--m[end];
int count = 0, result = 0;
for (auto& [k, v] : m)
{
count += v;
result = max(count, result);
}
return result;
}
};
/**
* Your MyCalendarThree object will be instantiated and called as such:
* MyCalendarThree* obj = new MyCalendarThree();
* int param_1 = obj->book(start,end);
*/
Java:
class MyCalendarThree {
private TreeMap<Integer, Integer> data;
public MyCalendarThree() {
data = new TreeMap<>();
}
public int book(int start, int end) {
data.put(start, data.getOrDefault(start, 0) + 1);
data.put(end, data.getOrDefault(end, 0) - 1);
int count = 0, result = 0;
for (int day : data.values()) {
count += day;
result = Math.max(result, count);
}
return result;
}
}
/**
* Your MyCalendarThree object will be instantiated and called as such:
* MyCalendarThree obj = new MyCalendarThree();
* int param_1 = obj.book(start,end);
*/
Python:
from sortedcontainers import SortedDict
class MyCalendarThree:
def __init__(self):
self.data = SortedDict()
def book(self, start: int, end: int) -> int:
self.data[start], self.data[end], count, result = self.data.get(start, 0) + 1, self.data.get(end, 0) - 1, 0, 0
for day in self.data.values():
count += day
result = max(result, count)
return result
# Your MyCalendarThree object will be instantiated and called as such:
# obj = MyCalendarThree()
# param_1 = obj.book(start,end)
