难度: Medium
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].
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)******
刚才我们是挨个比对,所以比对过程是O(N),但是由于我们已经保证了start按照升序排列,所以我们可以用二分算出合理的插入位置,然后看看是否有冲突即可
二分的实现可以用BST来实现,自己实现一下BST吧
beats 47.40%
class Node:
def __init__(self, start, end):
self.start = start
self.end = end
self.left = self.right = None
def insert(self, node):
if node.start >= self.end:
if not self.right:
self.right = node
return True
return self.right.insert(node)
elif node.end <= self.start:
if not self.left:
self.left = node
return True
return self.left.insert(node)
else:
return False
class MyCalendar(object):
def __init__(self):
self.root = None
def book(self, start, end):
"""
:type start: int
:type end: int
:rtype: bool
"""
if not self.root:
self.root = Node(start, end)
return True
return self.root.insert(Node(start, end))
思路 3 - 时间复杂度: O(lgN)- 空间复杂度: O(N)******
刚才我们是挨个比对,所以比对过程是O(N),但是由于我们已经保证了start按照升序排列,所以我们可以用二分算出合理的插入位置,然后看看是否有冲突即可
这次用最正宗的二分
- Return True if the insersion point is at the end of the array
- Return False if the insertion point overlaps with an existing calendar event
- the insertion point
right_idx
is odd, which means current events[right_idx-1](it's a start) < start < events[right_idx](it's a end) - end > events[right_idx])(it's a start), that means the current end ovelap with the events[right_idx](which is a start)
- the insertion point
- Otherwise return True
beats 100%
class MyCalendar(object):
def __init__(self):
self.events = []
def book(self, start, end):
"""
:type start: int
:type end: int
:rtype: bool
"""
right_idx = bisect.bisect_right(self.events, start)
if right_idx < len(self.events) and (right_idx & 1 or end > self.events[right_idx]):
return False
self.events[right_idx:right_idx] = [start, end]
return True