java - 调度3个元素的数组避免死锁
问题描述
给定一个共有 3 个 ["0","1","2"] 字符串元素的数组,所有初始呈现为可用,在分配 2 个元素之后,可以分配的最大元素数为 2(计划) ,不能再分配更多元素。基于这个前提,可能会出现以下场景。
(注:从代码中易于理解)
场景 1:进程 P1 请求一个元素,并被分配元素“0”(元素“0”然后变得无法分配给任何其他进程,因为它由进程 P1 持有),然后进程 P2 请求一个元素并被分配元素“1” - 现在阵列已达到其最大调度容量。任何进程(P1/P2)都可以在任何给定时间放弃它的元素。如果进程 P2 放弃它的元素“1”,那么进程 P3 请求一个元素,它应该被给予/分配元素“2”(因为元素 2 从未被分配过)。任何进程(P1/P3)都可以在任何给定时间放弃它的元素。如果进程 P1 放弃其元素“0”,则可以分配的下一个可用元素是元素“1”(因为数组中有两个元素(“1”、“0”)可用。
如何始终分配 2 个元素的最大值,避免元素饥饿?(假设进程以任何顺序都可以放弃它们的元素)
public class Main {
public static void main(String[] args)
{
myScheduler p = new myScheduler();
/* available elements: "0","1","2" imagine them being in a queue */
/* elements taken: non */
String show = p.requestElement(); //Should return "0"
/* available elements: "1","2" imagine them being in a queue */
/* elements taken: "0" */
System.out.println(show);
show = p.requestElement(); //Should return "1"
/* available elements: "2" queue cannot be empty */
/* elements taken: "0","1" imagine them being a linked list (at most 2 elements can be taken) */
System.out.println(show);
show = p.requestElement(); //Should return "full"
/* available elements: "2" queue cannot be empty */
/* elements taken: "0","1" note: it didn't change (at most 2 elements can be taken) */
System.out.println(show);
show = p.abandonElement("1"); //Should return "1"
/* available elements: "2","1" imagine them being in a queue */
/* elements taken: "0" */
System.out.println(show);
show = p.requestElement(); //Should return "2"
/* available elements: "1" queue cannot be empty */
/* elements taken: "0","2" (at most 2 elements can be taken) */
System.out.println(show);
show = p.requestElement(); //Should return "full"
/* available elements: "1" queue cannot be empty */
/* elements taken: "0","2" note: it did not change (at most 2 elements can be taken) */
System.out.println(show);
show = p.abandonElement("0"); //Should return "0"
/* available elements: "1","0" */
/* elements taken: "2" */
System.out.println(show);
show = p.requestElement(); //Should return "1"
/* available elements: "0" */
/* elements taken: "2","1" */
System.out.println(show);
}
public class myScheduler {
private String [] array;
public myScheduler()
{
array = {"0","1","2"};
}
public String requestElement(){
/*This function returns the element given to this request, if array has already assigned 2 elements, return "full" */
}
public String abandonElement(String element){
/*This function returns the element to be abandoned, if the element doesnt exist, return "not found" */
}
解决方案
所以这就是我要做的:构造一个LinkedList<>()
来跟踪最先预订的时间。(队列是 FIFO-先进先出)然后创建一个final ArrayList<>()
来跟踪您可以预订的时间。(您只希望您的用户能够预订0, 1,
和2
)
这是我的代码:
//Add the three choices. Make the ArrayList final so nobody can change it
public static final ArrayList<String> array = new ArrayList<>(); {{
array.add("0");
array.add("1");
array.add("2");
}}
public static LinkedList<String> linked = new LinkedList<>(); {{
linked.add("0");
linked.add("1");
linked.add("2");
}}
public String requestElement(){
/*This function returns the element given to this request, if array has already assigned 2 elements, return "full" */
String result = "full"; //Assume the result is full
if(linked.size()==1) { //If only one time is available return full
return result;
}
result = linked.poll(); //Poll the first item on the queue
return result;
}
public String abandonElement(String element){
/*This function returns the element to be abandoned, if the element doesnt exist, return "not found" */
String result = "Not Found";
if(array.contains(element)) { //If the arrayList contains the element, add it back to the LinkedList
linked.add(element);
result = element;
}
return result;
}
请让我知道,如果你有任何问题!
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