Employee Free Time

We are given a list scheduleof employees, which represents the working time for each employee.

Each employee has a list of non-overlappingIntervals, and these intervals are in sorted order.

Return the list of finite intervals representing common, positive-length free time forallemployees, 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 representingIntervalsin the form[x, y], the objects inside areIntervals, not lists or arrays. For example,schedule[0][0].start = 1, schedule[0][0].end = 2, andschedule[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.

Solution & Analysis

类似于Meeting Room问题

/**
 * 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>> avails) {
        List<Interval> result = new ArrayList<>();
        List<Interval> timeLine = new ArrayList<>();
        avails.forEach(e -> timeLine.addAll(e));
        Collections.sort(timeLine, ((a, b) -> a.start - b.start));

        Interval temp = timeLine.get(0);
        for (Interval each : timeLine) {
            if (temp.end < each.start) {
                result.add(new Interval(temp.end, each.start));
                temp = each;
            } else {
                temp = temp.end < each.end ? each : temp;
            }
        }
        return result;
    }
}

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