LeetCode 759. Employee Free Time

原题链接在这里：https://leetcode.com/problems/employee-free-time/description/

题目：

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:

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

题解：

k条已经排好序的链表. 利用minHeap进行merge. 

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.

Time Complexity: O(nlogk). k是employee的个数, schedule.size(). n是一共有多少interval.

Space: O(k).

AC Java:

/**
 * 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) {
        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;
    }
}