我需要编写一个简单的程序，该程序可以打印最多给定数字但不超过5秒的质数。经过一段时间后，是否可以使用某种计时器来中断方法？（但如果打印时间少于5秒，则不会中断）。提前致谢。
我的代码：
public class Primes {
private static boolean checkIfPrime(int x) {
if (x == 2) return true;
if (x % 2 == 0) return false;
int sqrt = (int) Math.sqrt(x) + 1;
for (int i = 3; i < sqrt; i = i + 2) if (x % i == 0) return false;
return true;
}
private static void printPrimesAndOperationTime(int n) {
long start = System.nanoTime();
for (int i = 2; i <= n; i++) if (checkIfPrime(i)) System.out.println(i);
long end = System.nanoTime();
long timeResult = end - start;
System.out.println("Printing time = " + timeResult + " [ns] => "
+ Math.round(timeResult * 100.0 / 1000000) / 100.0 + " [ms]");
}
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
printPrimesAndOperationTime(n);
}
}
使用Java并发API可以解决上述问题。请找到内联注释以进行代码遍历。
import java.util.Scanner;
import java.util.concurrent.*;
public class TimeoutInterval {
public static void main(String[] args) throws Exception {
ExecutorService executor = Executors.newSingleThreadExecutor(); // Start Single thread executor
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
Future future = executor.submit(new Primes(n)); // Find prime no.
try {
future.get(5, TimeUnit.SECONDS); // Set the time out of the prime no. search task
executor.shutdown();
} catch (TimeoutException e) {
executor.shutdown();
System.out.println("Terminated!");
}
executor.shutdownNow();
}
}
class Primes implements Runnable {
private final int number;
Primes(int number) {
this.number = number;
}
@Override
public void run() {
System.out.println("Started..");
printPrimesAndOperationTime(number);
System.out.println("Finished!");
}
private static boolean checkIfPrime(int x) {
if (x == 2) return true;
if (x % 2 == 0) return false;
int sqrt = (int) Math.sqrt(x) + 1;
for (int i = 3; i < sqrt; i = i + 2) if (x % i == 0) return false;
return true;
}
private static void printPrimesAndOperationTime(int n) {
long start = System.nanoTime();
for (int i = 2; i <= n && !Thread.interrupted(); i++) if (checkIfPrime(i)) {
System.out.println(i);
}
long end = System.nanoTime();
long timeResult = end - start;
System.out.println("Printing time = " + timeResult + " [ns] => "
+ Math.round(timeResult * 100.0 / 1000000) / 100.0 + " [ms]");
}
}
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