Description
Given an array of events where events[i] = [startDayi, endDayi]. Every event i starts at startDayi and ends at endDayi.
You can attend an event i at any day d where startTimei <= d <= endTimei. Notice that you can only attend one event at any time d.
Return the maximum number of events you can attend.
Example 1:
Input: events = [[1,2],[2,3],[3,4]]
Output: 3
Explanation: You can attend all the three events.
One way to attend them all is as shown.
Attend the first event on day 1.
Attend the second event on day 2.
Attend the third event on day 3.
Example 2:
Input: events= [[1,2],[2,3],[3,4],[1,2]]
Output: 4
Example 3:
Input: events = [[1,4],[4,4],[2,2],[3,4],[1,1]]
Output: 4
Example 4:
Input: events = [[1,100000]]
Output: 1
Example 5:
Input: events = [[1,1],[1,2],[1,3],[1,4],[1,5],[1,6],[1,7]]
Output: 7
Constraints:
- 1 <= events.length <= 10^5
- events[i].length == 2
- 1 <= startDayi <= endDayi <= 10^5
分析
题目的意思是：给若干区间段[si, ei]，每个单位时间 i 可且仅可完成一个任务，完成任务的要求为si <= i <= ei，问最多可完成多少个任务。
我参考了一下别人用堆的实现，首先对events进行排序，求出events的最大值n，然后从1遍历到n+1，对于遍历的i，找出以i开始的events，然后把endDayi加入堆中，然后去除end中小于i的数，剩下的数end大于等于i，如果非空，说明已经找到了一个符合要求的event，更新res。
代码
class Solution:
def maxEvents(self, events: List[List[int]]) -> int:
events.sort()
res=0
n=0
for i,j in events:
n=max(n,i,j)
end=[]
for i in range(1,n+1):
while(events and events[0][0]==i):
heapq.heappush(end,events.pop(0)[1])
while(end and end[0]<i):
heapq.heappop(end)
if(end):
heapq.heappop(end)
res+=1
return res