Employee Free Time

描述​

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.

(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.

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]]

Constraints:

• 1 <= schedule.length , schedule[i].length <= 50
• 0 <= schedule[i].start < schedule[i].end <= 10^8

代码​

// Employee Free Time// 时间复杂度O(n)，空间复杂度O(1)public class Solution {    public List<Interval> employeeFreeTime(List<List<Interval>> schedule) {        List<Interval> merged = merge(schedule.stream()                .flatMap(List::stream)                .collect(Collectors.toList()));        List<Interval> result = new ArrayList<>();        for (int i = 0; i < merged.size()-1; ++i) {            result.add(new Interval(merged.get(i).end, merged.get(i+1).start ));        }        return result;    }    private List<Interval> merge(List<Interval> intervals) {        List<Interval> result = new ArrayList<>();        for (int i = 0; i < intervals.size(); i++) {            insert(result, intervals.get(i));        }        return result;    }    private static List<Interval> insert(List<Interval> intervals,                                         Interval newInterval) {        for (int i = 0; i < intervals.size();) {            final Interval cur = intervals.get(i);            if (newInterval.end < cur.start) {                intervals.add(i, newInterval);                return intervals;            } else if (newInterval.start > cur.end) {                ++i;                continue;            } else {                newInterval.start = Math.min(newInterval.start, cur.start);                newInterval.end = Math.max(newInterval.end, cur.end);                intervals.remove(i);            }        }        intervals.add(newInterval);        return intervals;    }}