1. 程式人生 > >759. Employee Free Time

759. Employee Free Time

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, 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. Note: schedule and schedule[i] are lists with lengths in range [
1, 50]. 0 <= schedule[i].start < schedule[i].end <= 10^8. https://www.youtube.com/watch?v=VTgF52uGK0Y /** * Definition for an interval. * public class Interval { * int start; * int end; * Interval() { start = 0; end = 0; } * Interval(int s, int e) { start = s; end = e; } * } */ class Solution { public List<Interval> employeeFreeTime(List<List<Interval>> schedule) { // merge intervels and get the empty intervals List<Interval> res = new ArrayList<>(); List<Interval> time = new ArrayList<>(); for(List<Interval> list : schedule){ for(Interval interval : list){ time.add(interval); } } Collections.sort(time, (a, b) -> a.start - b.start); int end = time.get(0).end; for(Interval interval : time){ if(interval.start > end){ res.add(new Interval(end, interval.start)); // new Interval(end, interval.start), not new Interval<>() } end = Math.max(end, interval.end); } return res; } }