LeetCode 759 Employee Free Time

LeetCode

We are given a list schedule of employees, which represents the working time for each employee.

Each employee has a list of non-overlapping Intervals, and these intervals are in sorted order.

Return the list of finite intervals representing common, positive-length free time for all employees, also in sorted order.

(Even though we are representing Intervals in the form [x, y], the objects inside are Intervals, not lists or arrays. For example, schedule[0][0].start = 1, schedule[0][0].end = 2, and schedule[0][0][0] is not defined). Also, we wouldn’t include intervals like [5, 5] in our answer, as they have zero length.

Example 1:

Input: schedule = [[[1,2],[5,6]],[[1,3]],[[4,10]]]
Output: [[3,4]]
Explanation: There are a total of three employees, and all common
free time intervals would be [-inf, 1], [3, 4], [10, inf].
We discard any intervals that contain inf as they aren’t finite.

Example 2:

Input: schedule = [[[1,3],[6,7]],[[2,4]],[[2,5],[9,12]]]
Output: [[5,6],[7,9]]

Constraints:

1 <= schedule.length , schedule[i].length <= 50 0 <= schedule[i].start < schedule[i].end <= 10^8

题目大意：

给定员工的schedule，给所有员工的空余时间的区间

解题思路：

一开始考虑用Leetcode 056 merge intervals，然后求不能merge时的区间。后来想用Leetcode 253 meeting rooms II也是关于重合区间，只要active meetings为0时，就表示空余区间，此法更容易实现

解题步骤：

N/A

注意事项：

1. 用Leetcode 253 meeting rooms II的排序解法
2. 注意输入是每一个员工的schedule如[Interval(1, 2), Interval(5, 6)]，然后是所有员工schedule的列表

Python代码：

def test(self):
	TestCases.compare(self, self.employeeFreeTime([[Interval(1, 2), Interval(5, 6)], [Interval(1, 3)], [Interval(4, 10)]]), [Interval(3, 4)])

def employeeFreeTime(self, schedule: [[Interval]]) -> [Interval]:
	schedule_endpoints = []
	for li in schedule:
		for s in li:
			schedule_endpoints.append((s.start, 0)) # [(1, 0),(2,1),(5,0), (6,1)]
			schedule_endpoints.append((s.end, 1))
	schedule_endpoints.sort()
	res, free_time, active_schedule = [], [], 0
	for i in range(len(schedule_endpoints)):
		if schedule_endpoints[i][1] == 0:
			active_schedule += 1 # 0
		else:
			active_schedule -= 1 #
		if active_schedule == 0:
			free_time.append(schedule_endpoints[i][0]) #[2,5]
		if active_schedule == 1 and len(free_time) == 1:
			free_time.append(schedule_endpoints[i][0])
			free_interval = Interval(free_time[0], free_time[1])
			res.append(free_interval)
			free_time = []
	return res

算法分析：

时间复杂度为O(nlogn)，空间复杂度O(n)