FB面经prepare: task schedule II

followup是tasks是无序的.
一开始是有序的，比如说1, 1, 2, 1，一定要先执行第一个task1，然后等task1恢复，再执行第2个task1，再执行task2..... followup是无序的，就是不用按给的顺序执行，也就是可以先执行task1，然后task1还没恢复时，先执行task2, etc......

正确的做法应该是统计每个task的frequency，然后每次选frequency最高并且可以执行的task执行。

用maxHeap存每个task的剩余frequency

package TaskSchedule;
import java.util.*;

public class Solution2 {
    public class Element {
        int val;
        int appr;
        public Element(int value) {
            this.val = value;
            this.appr = 1;
        }
    }
    
    public int schedule(int[] arr, int recover) {
        HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
        PriorityQueue<Element> queue = new PriorityQueue<Element>(11, new Comparator<Element>() {
            public int compare(Element e1, Element e2) {
                return e2.appr-e1.appr;
            }
        });
        Element[] summary = new Element[10];
        for (int each : arr) {
            if (summary[each] == null) {
                summary[each] = new Element(each);
            }
            else summary[each].appr++;
        }
        for (Element elem : summary) {
            if (elem != null)
                queue.offer(elem);
        }
        
        
        
        
        int time = 0;
        LinkedList<Element> temp = new LinkedList<Element>();
        while (!queue.isEmpty() || !temp.isEmpty()) {
            if (!queue.isEmpty()) {
                Element cur = queue.poll();
                if (!map.containsKey(cur.val)) {
                    map.put(cur.val, time+recover+1);
                    cur.appr--;
                    if (cur.appr > 0) temp.offer(cur);
                    while (!temp.isEmpty()) {
                        queue.offer(temp.poll());
                    }
                    time++;
                }
                
                else { //map contains cur.val
                    if (time >= map.get(cur.val)) { //time is feasible
                        map.put(cur.val, time+recover+1);
                        cur.appr--;
                        if (cur.appr > 0) temp.offer(cur);
                        while (!temp.isEmpty()) {
                            queue.offer(temp.poll());
                        }
                        time++;
                    }
                    else { //time is not feasible
                        temp.offer(cur);
                    }
                }
            }
            
            else { //queue is empty, but temp is not empty
                while (!temp.isEmpty()) {
                    queue.offer(temp.poll());
                }
                time++;
            }
        }
        return time;
    }
    

    /**
     * @param args
     */
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        Solution2 sol = new Solution2();
        System.out.println(sol.schedule(new int[]{1, 1, 2, 2, 3, 3}, 2));
    }

}