我想创建一个比较器进行操作,使得到达时间较短的进程将首先出现在排序中,如果两个进程具有相同的到达时间,那么具有较低进程ID的进程将首先排序。我尝试了以下代码,但它似乎没有工作。有没有人看到它的缺陷?
public class FCFSComparator implements Comparator<Process>
{
public int compare(Process o1, Process o2)
{
int result = o1.getArrivalTime() - o2.getArrivalTime();
if(result == 0)
{
return (o1.getPid() < o2.getPid()) ? -1 : 1;
}
else
{
return result;
}
// return (result != 0 ? result : o1.getPid() - o2.getPid());
}
}
具体而言,给定以下流程
pid = 0 arrival time = 10
pid = 1 arrival time = 30
pid = 2 arrival time = 15
pid = 3 arrival time = 15
pid = 4 arrival time = 66
我在最后得到以下订购
Pid = 0 arrival time = 10
Pid = 2 arrival time = 15
Pid = 1 arrival time = 30
Pid = 4 arrival time = 66
Pid = 3 arrival time = 15
答案 0 :(得分:4)
我发现你的比较器没有任何问题。这是我的测试用例:
public class TestComparator {
static class Process {
int pid;
int arrivalTime;
Process(int pid, int arrivalTime) {
this.pid = pid;
this.arrivalTime = arrivalTime;
}
@Override
public String toString() {
return "Process [pid=" + pid + ", arrivalTime=" + arrivalTime + "]";
}
}
static class FCFSComparator implements Comparator<Process> {
public int compare(Process o1, Process o2) {
int result = o1.arrivalTime - o2.arrivalTime;
if (result == 0) {
return (o1.pid < o2.pid) ? -1 : 1;
} else {
return result;
}
}
}
public static void main(String[] args) {
List<Process> processes = Arrays.asList(
new Process(0, 10),
new Process(1, 30),
new Process(2, 15),
new Process(3, 15),
new Process(4, 66));
Collections.sort(processes, new FCFSComparator());
for (Process process : processes) {
System.out.println(process);
}
}
}
输出:
Process [pid=0, arrivalTime=10]
Process [pid=2, arrivalTime=15]
Process [pid=3, arrivalTime=15]
Process [pid=1, arrivalTime=30]
Process [pid=4, arrivalTime=66]
答案 1 :(得分:3)
我假设您正在比较的事情是int
。在两个变量相等的情况下,您仍然从内部比较返回1。这样的事情应该会有所帮助:
public class FCFSComparator implements Comparator<Process>
{
public int compare(Process o1, Process o2)
{
int result = o1.getArrivalTime() - o2.getArrivalTime();
if (result == 0)
{
return o1.getPid() - o2.getPid();
}
else
{
return result;
}
}
}
编辑:我检查了上面的代码,它确实输出了正确的顺序。我只能假设您的代码中有其他地方存在错误。
Pid = 0 arrival time = 10
Pid = 2 arrival time = 15
Pid = 3 arrival time = 15
Pid = 1 arrival time = 30
Pid = 4 arrival time = 66
完整的测试代码是:
import java.util.Comparator;
import java.util.List;
import java.util.ArrayList;
import java.util.Collections;
public class Main
{
public static void main(String[] args)
{
List<Process> processes = new ArrayList<Process>();
processes.add(new Process(10, 0));
processes.add(new Process(30, 1));
processes.add(new Process(15, 2));
processes.add(new Process(15, 3));
processes.add(new Process(66, 4));
Collections.sort(processes, new FCFSComparator());
for (Process process : processes)
System.out.println("Pid = " + process.getPid() + " arrival time = " + process.getArrivalTime());
}
static class FCFSComparator implements Comparator<Process>
{
public int compare(Process o1, Process o2)
{
int result = o1.getArrivalTime() - o2.getArrivalTime();
if (result == 0)
{
return o1.getPid() - o2.getPid();
}
else
{
return result;
}
}
}
static class Process
{
private int arrivalTime;
private int pid;
Process(int arrivalTime, int pid)
{
this.arrivalTime = arrivalTime;
this.pid = pid;
}
public int getArrivalTime()
{
return arrivalTime;
}
public int getPid()
{
return pid;
}
}
}
答案 2 :(得分:2)
我相信你想要这个:
public class FCFSComparator implements Comparator<Process> {
public int compare(Process o1, Process o2) {
if (o1.getArrivalTime() == o2.getArrivalTime()) {
return (o1.getPid() < o2.getPid()) ? -1 : 1;
}
return (o1.getArrivalTime() < o2.getArrivalTime()) ? -1 : 1;
}
}
答案 3 :(得分:1)
最简单的是:
public class FCFSComparator implements Comparator<Process>
{
public int compare(Process o1, Process o2)
{
int timeCmp = Integer.valueOf(o1.getArrivalTime()).compareTo(Integer.valueOf(o2.getArrivalTime()));
return (timeCmp != 0 ? timeCmp : Integer.valueOf(o1.getPid()).compareTo(Integer.valueOf(o2.getPid())));
}
}