759. Employee Free Time

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题目

  1. 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.

示例1:

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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.

示例2：

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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.

提示:

  1. schedule and schedule[i] are lists with lengths in range [1, 50].
  2. 0 <= schedule[i].start < schedule[i].end <= 10^8.

解法

解法一：

先把每个表头放在minHeap中. minHeap按照指向的Interval start排序.

poll出来的就是当前最小start的interval. 如果标记的时间比这个interval的start还小就说明出现了断裂也就是空余时间.

把标记时间增大到这个interval的end, 并且把这个interval所在链表的后一位加入minHeap中.

JAVA

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public class Interval {
		int start;
		int end;

		Interval() {
			start = 0;
			end = 0;
		}

		Interval(int s, int e) {
			start = s;
			end = e;
		}
	}

	public List<Interval> employeeFreeTime(List<List<Interval>> schedule) {
		List<Interval> res = new ArrayList<Interval>();
		PriorityQueue<Node> minHeap = new PriorityQueue<Node>(
				(a, b) -> schedule.get(a.employee).get(a.index).start - schedule.get(b.employee).get(b.index).start);

		int start = Integer.MAX_VALUE;
		for (int i = 0; i < schedule.size(); i++) {
			minHeap.add(new Node(i, 0));
			start = Math.min(start, schedule.get(i).get(0).start);
		}

		while (!minHeap.isEmpty()) {
			Node cur = minHeap.poll();
			if (start < schedule.get(cur.employee).get(cur.index).start) {
				res.add(new Interval(start, schedule.get(cur.employee).get(cur.index).start));
			}

			start = Math.max(start, schedule.get(cur.employee).get(cur.index).end);
			cur.index++;
			if (cur.index < schedule.get(cur.employee).size()) {
				minHeap.add(cur);
			}
		}

		return res;
	}

	class Node {
		int employee;
		int index;

		public Node(int employee, int index) {
			this.employee = employee;
			this.index = index;
		}
	}
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