FB面经prepare: task schedule II - neverlandly - 博客园

FB面经prepare: task schedule II - neverlandly - 博客园

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

 5     public class Element {
 6         int val;
 7         int appr;
 8         public Element(int value) {
 9             this.val = value;
10             this.appr = 1;
11         }
12     }
13     
14     public int schedule(int[] arr, int recover) {
15         HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
16         PriorityQueue<Element> queue = new PriorityQueue<Element>(11, new Comparator<Element>() {
17             public int compare(Element e1, Element e2) {
18                 return e2.appr-e1.appr;
19             }
20         });
21         Element[] summary = new Element[10];
22         for (int each : arr) {
23             if (summary[each] == null) {
24                 summary[each] = new Element(each);
25             }
26             else summary[each].appr++;
27         }
28         for (Element elem : summary) {
29             if (elem != null)
30                 queue.offer(elem);
31         }
32         
33         
34         
35         
36         int time = 0;
37         LinkedList<Element> temp = new LinkedList<Element>();
38         while (!queue.isEmpty() || !temp.isEmpty()) {
39             if (!queue.isEmpty()) {
40                 Element cur = queue.poll();
41                 if (!map.containsKey(cur.val)) {
42                     map.put(cur.val, time+recover+1);
43                     cur.appr--;
44                     if (cur.appr > 0) temp.offer(cur);
45                     while (!temp.isEmpty()) {
46                         queue.offer(temp.poll());
47                     }
48                     time++;
49                 }
50                 
51                 else { //map contains cur.val
52                     if (time >= map.get(cur.val)) { //time is feasible
53                         map.put(cur.val, time+recover+1);
54                         cur.appr--;
55                         if (cur.appr > 0) temp.offer(cur);
56                         while (!temp.isEmpty()) {
57                             queue.offer(temp.poll());
58                         }
59                         time++;
60                     }
61                     else { //time is not feasible
62                         temp.offer(cur);
63                     }
64                 }
65             }
66             
67             else { //queue is empty, but temp is not empty
68                 while (!temp.isEmpty()) {
69                     queue.offer(temp.poll());
70                 }
71                 time++;
72             }
73         }
74         return time;
75     }

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