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.

思路：这题是greedy的思想，首先按照end date排序，优先考虑课程结束早的课程，然后，每次加入一个课程，如果当前课程超过了end date，那么我们要剔除一门课，但是我们这里不是剔除刚加的那门，是剔除花费时间最长的一门课，因为我们的目标是尽量上的课的门数最多，那么我们要的就是要上更多的门，那么就是踢掉一门耗时最多的，说不定后面还可以加入更多的课；这里要学会java 8的lambda expression

class Solution {
    public int scheduleCourse(int[][] courses) {
        if(courses == null || courses.length == 0 || courses[0].length == 0) {
            return 0;
        }
        Arrays.sort(courses, (a, b) -> (a[1] - b[1]));
        PriorityQueue<Integer> pq = new PriorityQueue<Integer>((a, b) -> (b - a));
        int cur = 0;
        for(int i = 0; i < courses.length; i++) {
            cur += courses[i][0];
            pq.offer(courses[i][0]);
            if(cur > courses[i][1]) {
                cur -= pq.poll();
            }
        }
        return pq.size();
    }
}