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, 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.

 class Solution {
     public List<Interval> employeeFreeTime(List<List<Interval>> schedule) {
         List<Interval> res = new ArrayList<>();
         List<Interval> times = new ArrayList<>();
         for (List<Interval> list : schedule) {
             times.addAll(list);
         }
         Collections.sort(times, ((i1, i2) -> i1.start - i2.start));
         Interval pre = times.get();
         for (int i = ; 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;
     }
 }

其实我们可以不用sort,我们可以maintain一个k size的heap, k refers to the number of employees 然后每次从heap里面取一个，和之前的一个进行比较。