[LeetCode] 759. Employee Free Time

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

员工空闲时间。题意是给定员工的schedule列表，表示每个员工的工作时间。每个员工都有一个非重叠的时间段Intervals列表，这些时间段已经排好序。请返回表示所有员工的共同的，正数长度的空闲时间的列表，同样需要排好序。

思路还是扫描线算法。但是这里有一些细节需要解释一下，input给的是多个员工的schedule，以interval of intervals表示。有的员工的schedule是断开的，所以input里面会有一个员工多个interval的情况。这个题有两种思路做，一种是pq，一种是先排序再扫描。

pq的做法是，将所有人的所有interval加入一个以start time排序的pq，先弹出一个，记为temp，比较temp.end和pq.peek().start的大小关系。若两者没有交集，则发现了一个gap，将其加入结果集；若没有交集，则temp.end = Math.max(temp.end, pq.peek().end)。

时间O(nlogn)

空间O(n)

Java实现ONLY

class Solution {
    public List<Interval> employeeFreeTime(List<List<Interval>> schedule) {
        List<Interval> res = new ArrayList<>();
        PriorityQueue<Interval> pq = new PriorityQueue<>((a, b) -> a.start - b.start);
        schedule.forEach(e -> pq.addAll(e));
        Interval temp = pq.poll();
        while (!pq.isEmpty()) {
            if (temp.end < pq.peek().start) {
                res.add(new Interval(temp.end, pq.peek().start));
                temp = pq.poll();
            } else {
                temp = temp.end < pq.peek().end ? pq.peek() : temp;
                pq.poll();
            }
        }
        return res;
    }
}

扫描的做法非常类似。将所有人的所有interval放在一起，按start time排序。接着两两比较，看是否前一个interval.end < 后一个interval.start，若是则找到一个gap，加入结果集；若不是则接着往后找。

时间O(nlogn)

空间O(n)

Java实现

class Solution {
    public List<Interval> employeeFreeTime(List<List<Interval>> schedule) {
        List<Interval> res = new ArrayList<>();
        List<Interval> copy = new ArrayList<>();
        schedule.forEach(e -> copy.addAll(e));
        Collections.sort(copy, (a, b) -> a.start - b.start);
        Interval temp = copy.get(0);
        for (Interval each : copy) {
            if (temp.end < each.start) {
                res.add(new Interval(temp.end, each.start));
                temp = each;
            } else {
                temp = temp.end < each.end ? each : temp;
            }
        }
        return res;
    }
}

JavaScript实现

/**
 * @param {Interval[][]} schedule
 * @return {Interval[]}
 */
var employeeFreeTime = function(schedule) {
    let res = [];
    let copy = [];
    for (let employee of schedule) {
        for (let interval of employee) {
            copy.push(interval);
        }
    }
    copy.sort((a, b) => a.start - b.start);
    let temp = copy[0];
    for (each of copy) {
        if (temp.end < each.start) {
            res.push(new Interval(temp.end, each.start));
            temp = each;
        } else {
            temp = temp.end < each.end ? each : temp;
        }
    }
    return res;
};

扫描线相关题目

LeetCode 题目总结