Leetcode 759. Employee Free Time

Problem:

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.

Solution:

　　這道題簡單說說，Hard題中的easy題，index數組記錄了每一位員工正要處理的事物的索引，初始化爲0。用一個優先級隊列，隊列頂端是全部員工在處理的事物中開始時間最先的那個員工的編號和事務。用timestamp記錄了全部當前進行的事務中最遲的結束時間，若是下一個開始最先的事務的開始時間比當前最遲的結束時間還大，則推入結果集中，不然更新最遲的結束時間。

Code:

 

 1 /**  2  * Definition for an interval.  3  * struct Interval {  4  * int start;  5  * int end;  6  * Interval() : start(0), end(0) {}  7  * Interval(int s, int e) : start(s), end(e) {}  8  * };  9  */
10 class Solution { 11 public: 12     struct mycmp{ 13         bool operator()(pair<int,Interval> &p1,pair<int,Interval> &p2){ 14             return p1.second.start > p2.second.start; 15  } 16  }; 17     vector<Interval> employeeFreeTime(vector<vector<Interval>>& schedule) { 18         vector<Interval> result; 19         vector<int> index(schedule.size(),1); 20         priority_queue<pair<int,Interval>,vector<pair<int,Interval>>,mycmp> pq; 21         for(int i = 0;i != schedule.size();++i) 22             pq.push(make_pair(i,schedule[i][0])); 23         int timestamp = pq.top().second.end; 24         while(!pq.empty()){ 25             pair<int,Interval> top = pq.top(); 26             if(top.second.start > timestamp){ 27  Interval interval(timestamp,top.second.start); 28  result.push_back(interval); 29                 timestamp = top.second.end; 30  } 31             else{ 32                 timestamp = max(timestamp,top.second.end); 33  } 34  pq.pop(); 35             if(index[top.first] != schedule[top.first].size()){ 36  pq.push(make_pair(top.first,schedule[top.first][index[top.first]])); 37                 index[top.first]++; 38  } 39  } 40         return result; 41  } 42 };