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:
scheduleandschedule[i]are lists with lengths in range[1, 50].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;
}
}
