用Java实现的轮转调度算法

本文介绍了用Java实现的轮转调度算法的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

在braining小时，我终于崩溃了，纷纷前来结果是我不知道如何实现循环赛到Java。我尝试过不同的方法和最近我有......嗯，我举个例子来说。

AT =到达时间 BT =突发时间（执行时间）

起初，我有这样的行号（0,5; 6,9; 6,5; 15,10），其中位置的元素 0-2-4  再present到达时间和元素位置 1-3-5 重present破裂时间。  我的code是到目前为止，这个输入变成一类，称为Process附带了一个  构造函数：进程（字符串名称，诠释AT，INT BT）。我已经与在的ArrayList 分离过程。  所以，现在我有一个 ArrayList的alst = [P0，P1，P2，P3]，其中P0有在0和BT 5等 ..

我创建了一个将返回我已经切断与时间的量子进程列表的方法 - 例如如果的（0,5; 6,9; 6,5; 15,10）我会得到一个结果： [P0， P0，P1，P1，P1，P2，P2，P3，P3，P3，P3]

因此​​循环赛的方法是一种方法，其中每一道工序得到的量子的时间，我已经选择3执行。

P0与AT 0安培; BT 3进来 - 添加至最终名单（时间的流逝 = 3）

P0与AT 0安培; BT 2进来 - 添加至最终名单（时间的流逝 = 5）

P0完成

P1与AT 6安培; BT 3进来 - 添加至最终名单（时间的流逝 = 9）

下一个P1被添加到队列

P2与AT 6安培; BT 3进来 - 添加至最终名单（时间的流逝 = 12）

下一个P2被添加到队列

P1与AT 6安培; BT 3进来的队列 - 添加至最终名单 （时间的推移= 15）

在接下来的P1进入队列

P3到达时，加在最终名单（时间过去了= 18）

P1来自队列 - 添加到最终名单

而这正是我觉得我的心一点已经崩溃了，我不知道怎么排队它们。

结果应该是这样的： [P0，P0，P1，P2，P1，P3，P2，P1，P3，P3，P3]

编辑：我的$基础上给出的第一个答案C $ CD。仍然无法正常工作。

公开的ArrayList roundRobinJarjestus（ArrayList中pstlst）{     ArrayList的队列=新的ArrayList（）; //järjekord，algusestühi     ArrayList的uuspst =新的ArrayList（）; //     queue.add（pstlst.get（0））;     INT I = 0;     双时间= 0;     双pworkTime = 0;     INT KVANT = 3;     而（ⅰ＆其中; pstlst.size（））{         Protsess p值=（Pr​​otsess）queue.get（ⅰ）; //第一个过程是采取         pworkTime = p.getTooaeg（）; //执行时间         时间=时间+ pworkTime;         如果（（（Protsess）pstlst.get第（i + 1））getSaabumisaeg（）＆所述;时间）{//如果下一个到达时间比过去的时间低             queue.add（pstlst.get第（i + 1））;         }         如果（pworkTime-KVANT大于0）{//如果烹饪时间 - 量子大于零，仍然留下一些东西来执行             p.setTooaeg（pworkTime-KVANT）;             queue.add（对）;         }         uuspst.add（queue.get（ⅰ））;         I = I + 1;     }     返回uuspst; }

解决方案

您可以保持等待处理的队列，并使用下面的算法：

接第一处理在队列中（如果它不是空的）。将它添加到输出列表。

执行它的时间一给定的量子（或更少，如果剩余时间是小于一个量子），并从该过程的剩余时间减去该量子

如果新的流程已经抵达，将它们添加到队列的末尾。

如果最后执行的处理没有结束（即，它的剩余时间不为0），将其添加到等待队列的末尾。

转到步骤1，如果有剩下的进程。

After hours of braining I've finally crashed and have come to result that I have no clue how to implement round robin into java. I've tried different approaches and the closest I've got.. well i explain with an example..

AT = Arrival Time BT = Burst Time (Execution Time)

At first i have this row of numbers (0,5;6,9;6,5;15,10) where elements in position 0-2-4 represent arrival times and elements in position 1-3-5 represent burst times. My code is so far that this input is turned into a class, called Process which comes with a constructor: Process(String name, int AT, int BT). I've separated processes with in the ArrayList. So now i have an ArrayList alst = [P0,P1,P2,P3] where P0 has AT 0 and BT 5 and so on..

I created a method which will return me a list of processes which have been cut with a quantum of time - For example in case of (0,5;6,9;6,5;15,10) i will get a result: [P0,P0,P1,P1,P1,P2,P2,P3,P3,P3,P3]

So round robin method is a method where every process gets quantum time for execution which I've chosen 3.

P0 with AT 0 & BT 3 comes in - added to the final list (time passed = 3)

P0 with AT 0 & BT 2 comes in - added to the final list (time passed = 5)

P0 finished

P1 with AT 6 & BT 3 comes in - added to the final list (time passed = 9)

next P1 is added to the queue

P2 with AT 6 & BT 3 comes in - added to the final list (time passed = 12)

next P2 is added to the queue

P1 with AT 6 & BT 3 comes in from queue - added to the final list (time passed = 15)

next P1 goes to the queue

P3 arrives, added to the final list (time passed = 18)

P1 comes from the queue - added to the final list

And that's the point where I feel like my mind has crashed and i have no idea how to queue them.

The result should look like: [P0,P0,P1,P2,P1,P3,P2,P1,P3,P3,P3]

EDIT: what I coded based on the first answer given. Still doesn't work..

public ArrayList roundRobinJarjestus(ArrayList pstlst){ ArrayList queue = new ArrayList();// järjekord, alguses tühi ArrayList uuspst = new ArrayList(); // queue.add(pstlst.get(0)); int i = 0; double time = 0; double pworkTime = 0; int kvant = 3; while (i < pstlst.size()){ Protsess p = (Protsess) queue.get(i); //first process is taken pworkTime = p.getTooaeg(); //execute time time = time + pworkTime; if (((Protsess) pstlst.get(i+1)).getSaabumisaeg() < time){ //if next arrival time is lower than time passed queue.add(pstlst.get(i+1)); } if (pworkTime-kvant > 0){ //if worktime - quantum is higher than zero and still left something to execute p.setTooaeg(pworkTime-kvant); queue.add(p); } uuspst.add(queue.get(i)); i = i+1; } return uuspst; }

解决方案

You can maintain a queue of waiting processes and use the following algorithm:

Pick the first process in the queue(if it is not empty). Add it to the output list.

Execute it for a given quantum of time(or less if its remaining time is less then one quantum) and subtract this quantum from the remaining time of this process.

If new processes have arrived, add them to the end of the queue.

If the last executed process is not finished(that is, its remaining time is not 0), add it to the end of the waiting queue.

Go to step 1 if there are any processes left.