630. Course Schedule III

There are n different online courses numbered from 1 to n. You are given an array courses where courses[i] = [durationi, lastDayi] indicate that the ith course should be taken continuously for durationi days and must be finished before or on lastDayi.

You will start on the 1st day and you cannot take two or more courses simultaneously.

Return the maximum number of courses that you can take.

Example 1:

Input: courses = [[100,200],[200,1300],[1000,1250],[2000,3200]]
Output: 3
Explanation: 
There are totally 4 courses, but you can take 3 courses at most:
First, take the 1st course, it costs 100 days so you will finish it on the 100th day, and ready to take the next course on the 101st day.
Second, take the 3rd course, it costs 1000 days so you will finish it on the 1100th day, and ready to take the next course on the 1101st day. 
Third, take the 2nd course, it costs 200 days so you will finish it on the 1300th day. 
The 4th course cannot be taken now, since you will finish it on the 3300th day, which exceeds the closed date.

Example 2:

Input: courses = [[1,2]]
Output: 1

Example 3:

Input: courses = [[3,2],[4,3]]
Output: 0

Constraints:

• 1 <= courses.length <= 104
• 1 <= durationi, lastDayi <= 104

public class Solution {
    public int scheduleCourse(int[][] courses) {
        Arrays.sort(courses,(a,b)->a[1]-b[1]); //Sort the courses by their deadlines (Greedy! We have to deal with courses with early deadlines first)
        PriorityQueue<Integer> pq=new PriorityQueue<>((a,b)->b-a);
        int time=0;
        for (int[] c:courses) 
        {
            time+=c[0]; // add current course to a priority queue
            pq.add(c[0]);
            if (time>c[1]) time-=pq.poll(); //If time exceeds, drop the previous course which costs the most time. (That must be the best choice!)
        }        
        return pq.size();
    }
}

标签：630,course,time,will,Course,take,courses,III,day
来源： https://www.cnblogs.com/wentiliangkaihua/p/14726938.html