EmployeeFreeTime.java

/** 

http://206.81.6.248:12306/leetcode/employee-free-time/description

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.
*/

public EmployeeFreeTime {
	public List<Interval> employeeFreeTime(List<List<Interval>> avails) {
		assert(avails != null);
		List<Interval> res = new ArrayList();
		if (avails.size() == 0) return res;
		List<Interval> intervals = new ArrayList();
		avails.forEach(a -> intervals.addAll(a));
		Collections.sort(intervals, (i1, i2) -> (i1.start - i2.start));
		Interval cur = intervals.get(0);
		for (int i = 1; i < intervals.size(); i++) {
			Interval next = intervals.get(i);
			if (cur.end < next.start) {
				res.add(new Interval(cur.end, next.start));
			}
			cur = next.end > cur.end ? next : cur;
		}
		return res;
	}
}