0630. Course Schedule III (H)

Course Schedule III (H)

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 <= 10^4
• 1 <= durationi, lastDayi <= 10^4

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题意

给定一系列课程的截止时间和持续时间，问最多能上多少门课。

思路

贪心。先按照截止时间从小到大排列，遍历课程，如果上完当前课程不会超时，则选择上这门课，并将当前课程的时长加入大顶堆中；如果上完当前课程会超时，那么查看堆顶时长是否比当前课程时长大，若大则取出堆顶课程，将当前课程时长入堆，进行替换。最终得到的堆的size即是答案。

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代码实现

Java

‘class Solution {
    public int scheduleCourse(int[][] courses) {
        Arrays.sort(courses, (a, b) -> a[1] - b[1]);
        Queue<Integer> pq = new PriorityQueue<>((a, b) -> b - a);
        int time = 0;

        for (int[] course : courses) {
            if (time + course[0] <= course[1]) {
                time += course[0];
                pq.offer(course[0]);
            } else if (!pq.isEmpty() && pq.peek() > course[0]) {
                time -= pq.poll();
                time += course[0];
                pq.offer(course[0]);
            }
        }

        return pq.size();
    }
}