759. Employee Free Time

759. Employee Free Time

Hard

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:

1. schedule and schedule[i] are lists with lengths in range [1, 50].
2. 0 <= schedule[i].start < schedule[i].end <= 10^8.

----------------------
不用想太复杂
把所有的interval都插入到priorityQueue，然后找gap。题目中给的多少个人其实没有什么意义
O(n), n为总的interval数量

/**
 * 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) {
        PriorityQueue<Interval> que = new PriorityQueue<>((a, b) -> a.start - b.start);
        
        for (List<Interval> list : schedule) {
            for (Interval i : list) {
                que.add(i);
            }
        }
        
        List<Interval> rt = new ArrayList<>();
        int max = -1;
        while (!que.isEmpty()) {
            Interval top = que.poll();
            if (max != -1 && top.start > max) {
                rt.add(new Interval(max, top.start));
            }
            max = Math.max(max, top.end);
        }
        
        return rt;
    }
}