Problem
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.
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]]
(Even though we are representing Intervals in the form [x, y], the objects inside are Intervals, not lists or arrays. For example, schedule0.start = 1, schedule0.end = 2, and schedule0[0] is not defined.)
Also, we wouldn"t include intervals like [5, 5] in our answer, as they have zero length.
Note:
schedule and schedule[i] are lists with lengths in range [1, 50].
0 <= schedule[i].start < schedule[i].end <= 10^8.
//put all intervals together, coz eventually we need //non-overlapping intervals from everyone //Can use a PriorityQueue to sort all intervals by start //or just use List and apply Collections.sort() //several formats for customizing comparator //https://www.mkyong.com/java8/java-8-lambda-comparator-example/ //after sorted, iterate the sorted list //if (pre.end < cur.start) --> save new Interval(pre.end, cur.start) class Solution { public ListemployeeFreeTime(List > schedule) { List
res = new ArrayList<>(); List times = new ArrayList<>(); for (List list: schedule) { times.addAll(list); } Collections.sort(times, ((i1, i2)->i1.start-i2.start)); Interval pre = times.get(0); for (int i = 1; i < times.size(); i++) { Interval cur = times.get(i); if (cur.start <= pre.end) { pre.end = cur.end > pre.end ? cur.end : pre.end; } else { res.add(new Interval(pre.end, cur.start)); pre = cur; } } return res; } }
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