729. My Calendar I

729. My Calendar I

难度: Medium

刷题内容

原题连接

内容描述

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

Your class will have the 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 double booking happens when two events have some non-empty intersection (ie., there is some time that is common to both events.)

For each call to the method MyCalendar.book, return true if the event can be added to the calendar successfully without causing a double 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(15, 25); // returns false
MyCalendar.book(20, 30); // returns true
Explanation: 
The first event can be booked.  The second can't because time 15 is already booked by another event.
The third event can be booked, as the first event takes every time less than 20, but not including 20.
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].

解题方案

思路 1
- 时间复杂度: O(N^2)- 空间复杂度: O(N)

按照start的顺序排列插入events,然后挨个比对,如果重叠了就return False,如果没有就把当前events插入到正确的排列顺序后,return True即可

beats 24.71%

class MyCalendar(object):

    def __init__(self):
        self.events = []

    def book(self, start, end):
        """
        :type start: int
        :type end: int
        :rtype: bool
        """
        for i in range(len(self.events)):
            time = self.events[i]
            if start < time[1] and end > time[0]:
                return False
            # if start >= time[1]:
            if start < time[0]:
                self.events.insert(i, (start, end))
                return True
        self.events.append((start, end))
        return True

思路 2
- 时间复杂度: O(lgN)- 空间复杂度: O(N)

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