贪心算法（1）

原题：

/** * Created by gouthamvidyapradhan on 27/06/2017. * <p> * 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. * <p> * Given n online courses represented by pairs (t,d), your task is to find the maximal number of courses that can be taken. * <p> * 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. * <p> * Note: * The integer 1 <= d, t, n <= 10,000. * You can't take two courses simultaneously. * <p> * Solution: O(N log N) * 1. Sort the courses with earliest deadline time (Greedy sort) * 2. Maintain a max-heap of course duration. * 3. Iterate through each course and increment the total time by current course time and include this in the * max-heap created in step 2. * 4. If the total time exceeds the current course deadline then, remove the course with max duration from max-heap * inorder to accommodate the new course. */

答案：

public class CourseScheduleIII {    public static void main(String[] args) throws Exception {        int[][] course = {{5, 5}, {2, 6}, {4, 6}};        System.out.println(new CourseScheduleIII().scheduleCourse(course));    }    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) {            time += course[0];            pq.add(course[0]);            if (time > course[1])                time -= pq.poll();        }        return pq.size();    }}