双端队列实现串行处理实现并发

ArrayDeque<Runnable> 通过try和finally机制，我们将传入的Runnable对象重新封装成一个新的Runnable对象，在新的Runnable的run方法的try块中执行Runnable的run方法，在finally中调用执行队列头部Runnable对象出队列，并放入线程池执行的方法

import java.util.ArrayDeque;
import java.util.concurrent.BlockingQueue;
import java.util.concurrent.LinkedBlockingDeque;
import java.util.concurrent.ThreadFactory;
import java.util.concurrent.ThreadPoolExecutor;
import java.util.concurrent.TimeUnit;
import java.util.concurrent.atomic.AtomicInteger;

public class SerialExecutor {
private Runnable mActive;
private ArrayDeque<Runnable> mArrayDeque = new ArrayDeque<>();

private static final int CPU_COUNT = Runtime.getRuntime().availableProcessors();
private static final int CORE_POOL_SIZE = CPU_COUNT + 1;
private static final int MAXIMUM_POOL_SIZE = CPU_COUNT * 2 + 1;
private static final int KEEP_ALIVE = 1;
private static final BlockingQueue<Runnable> sPoolWorkQueue =
new LinkedBlockingDeque<>(128);
private static final ThreadFactory sThreadFactory = new ThreadFactory() {
private final AtomicInteger mCount = new AtomicInteger(1);
@Override
public Thread newThread(Runnable r) {
return new Thread(r, "Serial thread #" + mCount.getAndIncrement());
}
};
private static final ThreadPoolExecutor THREAD_EXECUTOR = new ThreadPoolExecutor(CORE_POOL_SIZE,
MAXIMUM_POOL_SIZE, KEEP_ALIVE, TimeUnit.SECONDS, sPoolWorkQueue, sThreadFactory);

public synchronized void execute(final Runnable r) {
mArrayDeque.offer(new Runnable() {
@Override
public void run() {
try {
r.run();
} finally {
scheduleNext();
}
}
});
// 第一次入队列时mActivie为空，因此需要手动调用scheduleNext方法
if (mActive == null) {
scheduleNext();
}
}

private void scheduleNext() {
if ((mActive = mArrayDeque.poll()) != null) {
THREAD_EXECUTOR.execute(mActive);
}
}

public static void main(String[] args) {
SerialExecutor serialExecutor = new SerialExecutor();
for (int i = 0; i < 10; i ++) {
final int j = i;
serialExecutor.execute(new Runnable() {
@Override
public void run() {
System.out.println("The num is :" + (j + 1));
try {
Thread.sleep(1000);
} catch (InterruptedException e) {
e.printStackTrace();
}
}
});
}
}
}