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));
     }

 }