1：有一个总任务A，分解为子任务A1 A2 A3 ...，任何一个子任务失败后要快速取消所有任务，请写程序模拟。
「请寻求最优解，不要只是粗暴wait（）」
本题解题思路：Fork/Join
通常使用其更专门的类型之一 RecursiveTask（可以返回结果）或 RecursiveAction。
Oracle 官方文档：https://docs.oracle.com/javase/tutorial/essential/concurrency/forkjoin.html
主功能就是将一个大任务拆分成多个小任务进行处理。
处理过程中只要有个小任务失败报错，剩下的任务将可能被立即停止。
以下为代码实现：
1.1：实现代码
public class SonTask extends RecursiveAction {
private static final Logger logger = LoggerFactory.getLogger(SonTask.class);
/**
* 总共任务量
**/
private final int taskCount;
/**
* 当前task被分配的任务量
**/
private final int taskMete;
/**
* 当前task序号
**/
private int taskRank;
/**
* 每个task最大可处理任务量
**/
private final int maxTask = 1;
public SonTask(int taskCount) {
this.taskCount = taskCount;
this.taskMete = taskCount;
}
private SonTask(int taskCount, int taskMete, int taskRank) {
this.taskCount = taskCount;
this.taskMete = taskMete;
this.taskRank = taskRank;
}
@Override
protected void compute() {
// 任务分配量是否满足处理条件，不满足则将任务再拆分
if (taskMete == maxTask) {
printSelf();
} else {
List<SonTask> sonTaskList = new ArrayList<>();
for (int i = 1; i <= taskCount; i++) {
sonTaskList.add(new SonTask(taskCount, 1, i));
}
// 执行所有任务
invokeAll(sonTaskList);
}
}
/**
* task 1 正常结束 ->
* task 2 执行报错 ->
* task 3 直接终止
**/
private void printSelf() {
logger.info("SON TASK RANK [{}] START", taskRank);
try {
TimeUnit.SECONDS.sleep(taskRank * 3);
if (taskRank == 2) {
logger.error("eroor occured");
throw new RuntimeException("error");
} else {
logger.info("TASK [{}] OVER", taskRank);
}
} catch (InterruptedException e) {
e.printStackTrace();
}
}
}
1.2：测试
public class StartMain {
public static void main(String[] args) {
ForkJoinPool pool = new ForkJoinPool(10);
SonTask sonTask = new SonTask(3);
pool.invoke(sonTask);
}
}
在task 1结束后由于task 2报错了，task 3被取消执行。
看看ForkJoinTask#invokeAll(Collection
If any task encounters an exception, others may be cancelled.
/**
* Forks all tasks in the specified collection, returning when
* {@code isDone} holds for each task or an (unchecked) exception
* is encountered, in which case the exception is rethrown. If
* more than one task encounters an exception, then this method
* throws any one of these exceptions. If any task encounters an
* exception, others may be cancelled. However, the execution
* status of individual tasks is not guaranteed upon exceptional
* return. The status of each task may be obtained using {@link
* #getException()} and related methods to check if they have been
* cancelled, completed normally or exceptionally, or left
* unprocessed.
*
* @param tasks the collection of tasks
* @param <T> the type of the values returned from the tasks
* @return the tasks argument, to simplify usage
* @throws NullPointerException if tasks or any element are null
*/
public static <T extends ForkJoinTask<?>> Collection<T> invokeAll(Collection<T> tasks) {
...
}
2：请用两个线程交替输出A1B2C3D4...，A线程输出字母，B线程输出数字，要求A线程首先执行，B线程其次执行！（多种同步机制的运用）
「请寻求最优解，不要简单的synchronized」
本题解题思路：ReentrantLock、Condtional
1：利用Conditon#await、Condition#signal 进行线程之间的通信，替代Object#wait、Object#notify。
2：勿使用Thread#join 这种阻塞主线程的方式，也达不到该题的需求。
Condition的类注释：
/**
* {@code Condition} factors out the {@code Object} monitor
* methods ({@link Object#wait() wait}, {@link Object#notify notify}
* and {@link Object#notifyAll notifyAll}) into distinct objects to
* give the effect of having multiple wait-sets per object, by
* combining them with the use of arbitrary {@link Lock} implementations.
* Where a {@code Lock} replaces the use of {@code synchronized} methods
* and statements, a {@code Condition} replaces the use of the Object
* monitor methods.
* ...
*/
2.1：实现代码
public class StartMain {
private static final Logger logger = LoggerFactory.getLogger(StartMain.class);
private static final ReentrantLock lock = new ReentrantLock();
private static final String[] arr = new String[]{"A1", "B2", "C3", "D4"};
private static final AtomicInteger index = new AtomicInteger(0);
public static void main(String[] args) {
Condition conditionA = lock.newCondition();
Condition conditionB = lock.newCondition();
Thread threadA = new Thread(() -> {
while (index.get() < arr.length) {
try {
lock.lock();
logger.info(arr[index.get()]);
index.incrementAndGet();
conditionB.signal();
conditionA.await();
} catch (Exception e) {
e.printStackTrace();
} finally {
lock.unlock();
}
}
}, "thread-A");
Thread threadB = new Thread(() -> {
while (index.get() < arr.length) {
try {
lock.lock();
conditionB.await();
logger.info(arr[index.get()]);
index.incrementAndGet();
conditionA.signal();
} catch (Exception e) {
e.printStackTrace();
} finally {
lock.unlock();
}
}
}, "thread-B");
threadB.start();
// 为了使测试更加逼真，先让B开始
try {
TimeUnit.SECONDS.sleep(2);
} catch (InterruptedException e) {
e.printStackTrace();
}
threadA.start();
}
}
2.2：Condition#await
/**
* Causes the current thread to wait until it is signalled or
* {@linkplain Thread#interrupt interrupted}.
*
* <p>The lock associated with this {@code Condition} is atomically
* released and the current thread becomes disabled for thread scheduling
* purposes and lies dormant until <em>one</em> of four things happens:
*
* ...
* @throws InterruptedException if the current thread is interrupted
* (and interruption of thread suspension is supported)
*/
void await() throws InterruptedException;
**The lock associated with this {@code Condition} is atomically released and the current thread becomes disabled **
for thread scheduling purposes and lies dormant until ...
调用了Condition.await()方法后该线程所持有的Lock锁会被释放掉，并且当前线程会变得不可用（阻塞），直到调用了Condtion.signal()方法。
3：华为面试题
「请寻求最优解，不要简单的生产者-消费者模式」
有一个生产奶酪的厂家，每天需要生产100000份奶酪卖给超市，通过一辆货车发货，送货车每次送100份。
厂家有一个容量为1000份的冷库，用于奶酪保鲜，生产的奶酪需要先存放在冷库，运输车辆从冷库取货。
厂家有三条生产线，分别是牛奶供应生产线，发酵剂制作生产线，奶酪生产线。
生产每份奶酪需要2份牛奶和一份发酵剂。
请设计生产系统？
本题解题思路： BlockingDeque阻塞队列、Atomic原子类
Oracle 官方文档：https://docs.oracle.com/javase/7/docs/api/java/util/concurrent/BlockingDeque.html
3.1：api注释
1：BlockingDeque#take()
/**
* Retrieves and removes the head of the queue represented by this deque
* (in other words, the first element of this deque), waiting if
* necessary until an element becomes available.
*
* <p>This method is equivalent to {@link #takeFirst() takeFirst}.
*
* @return the head of this deque
* @throws InterruptedException if interrupted while waiting
*/
E take() throws InterruptedException;
该方法若从队列中取不到元素会造成当前线程阻塞，直到拿到元素为止。
2：BlockingDeque#put(E e)
/**
* Inserts the specified element into the queue represented by this deque
* (in other words, at the tail of this deque), waiting if necessary for
* space to become available.
*
* <p>This method is equivalent to {@link #putLast(Object) putLast}.
*
* @param e the element to add
* @throws InterruptedException {@inheritDoc}
* @throws ClassCastException if the class of the specified element
* prevents it from being added to this deque
* @throws NullPointerException if the specified element is null
* @throws IllegalArgumentException if some property of the specified
* element prevents it from being added to this deque
*/
void put(E e) throws InterruptedException;
当队列容量达到上限时，其他元素无法入队且使当前线程阻塞，直到队有可用空间为止。
3.2：实现代码
public class ProductOnlineBus {
private static final Logger logger = LoggerFactory.getLogger(ProductOnlineBus.class);
/**
* 生产奶酪数量
**/
private final int prodNum;
/**
* 牛奶=奶酪*2
**/
private final int milkMultiple = 2;
/**
* 发酵剂=奶酪*1
**/
private final int fjjMultiple = 1;
/**
* 奶酪仓库容量
**/
private final int cheeseCapacity = 1000;
/**
* 单词运输奶酪数量
**/
private final int truckCapacity = 100;
/**
* 总共需要运输多少次
**/
private final int needTruckTimes;
/**
* 生产线--阻塞队列
**/
private final BlockingDeque<MiikNode> milkNodeBlockingDeque;
private final BlockingDeque<FJJNode> fjjNodeBlockingDeque;
private final BlockingDeque<CheeseNode> cheeseNodeBlockingDeque;
/**
* 实际运输次数
**/
private final AtomicInteger trucked = new AtomicInteger(0);
/**
* 各生产线生产次数
**/
private final AtomicInteger milkProded = new AtomicInteger(0);
private final AtomicInteger fjjProded = new AtomicInteger(0);
private final AtomicInteger cheeseProded = new AtomicInteger(0);
public ProductOnlineBus(int prodNum) {
if ((prodNum % truckCapacity) != 0) {
throw new RuntimeException("请输入truckCapacity的倍数");
}
this.prodNum = prodNum;
this.milkNodeBlockingDeque = new LinkedBlockingDeque<>(milkMultiple);
this.fjjNodeBlockingDeque = new LinkedBlockingDeque<>(fjjMultiple);
this.cheeseNodeBlockingDeque = new LinkedBlockingDeque<>(cheeseCapacity);
this.needTruckTimes = prodNum / truckCapacity;
}
public void starter() {
new Thread(() -> {
int len = prodNum * milkMultiple;
for (int i = 0; i < len; i++) {
try {
milkNodeBlockingDeque.put(new MiikNode(i));
milkProded.incrementAndGet();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
}, "MilkThread").start();
new Thread(() -> {
int len = prodNum * fjjMultiple;
for (int i = 0; i < len; i++) {
try {
fjjNodeBlockingDeque.put(new FJJNode(i));
fjjProded.incrementAndGet();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
}, "FJJThread").start();
new Thread(() -> {
for (int i = 0; i < prodNum; i++) {
try {
for (int j = 0; j < milkMultiple; j++) {
milkNodeBlockingDeque.take();
}
for (int j = 0; j < fjjMultiple; j++) {
fjjNodeBlockingDeque.take();
}
cheeseNodeBlockingDeque.put(new CheeseNode(i));
cheeseProded.incrementAndGet();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
}, "CheeseThread").start();
new Thread(() -> {
while (trucked.get() < needTruckTimes) {
try {
for (int i = 0; i < truckCapacity; i++) {
cheeseNodeBlockingDeque.take();
}
trucked.incrementAndGet();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
logger.info("over of->cheese:[{}],milk:[{}],fjj[{}],truck:[{}]",
cheeseProded.get(), milkProded.get(), fjjProded.get(), trucked.get());
}, "TruckThread").start();
}
/**
* 牛奶
**/
private class MiikNode {
public MiikNode(int seq) {
logger.info("生产牛奶[{}]...", seq);
}
}
/**
* 发酵剂
**/
private class FJJNode {
public FJJNode(int seq) {
logger.info("生产发酵剂[{}]...", seq);
}
}
/**
* 奶酪
**/
private class CheeseNode {
public CheeseNode(int seq) {
logger.info("生产奶酪[{}]...", seq);
}
}
}
3.3：运行
public class StartMain {
public static void main(String[] args) {
ProductOnlineBus pb = new ProductOnlineBus(100000);
pb.starter();
}
}
