题目描述：

LeetCode 630. Course Schedule III

There are n different online courses numbered from 1 to n. Each course has some duration(course length) t and closed on dth day. A course should be taken continuously for t days and must be finished before or on the dth day. You will start at the 1st day.

Given n online courses represented by pairs (t,d), your task is to find the maximal number of courses that can be taken.

Example:

Input: [[100, 200], [200, 1300], [1000, 1250], [2000, 3200]]
Output: 3
Explanation: 
There're 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.

Note:

1. The integer 1 <= d, t, n <= 10,000.
2. You can't take two courses simultaneously.

题目大意：

给定n门课程，每门课程用二元组（时长，最迟完成时间）表示

求最多可以选择多少门课程

解题思路：

贪心算法（Greedy Algorithm）

课程根据最迟完成时间从小到大排序

遍历课程，利用优先队列（时长最大堆）维护当前满足最迟完成时间约束的课程时长，弹出不满足约束条件的课程时长

返回优先队列的长度

Java代码：

public class Solution {
    public int scheduleCourse(int[][] courses) {
        Arrays.sort(courses, new Comparator<int[]>(){
            public int compare(int[] a, int[] b) {
                return a[1] - b[1];
            }
        });
        PriorityQueue<Integer> pq = new PriorityQueue<>(Comparator.reverseOrder());
        int sum = 0;
        for (int course[] : courses) {
            int d = course[0], e = course[1];
            pq.add(d);
            sum += d;
            while (sum > e) {
                sum -= pq.poll();
            }
        }
        return pq.size();
    }
}

 

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