三個(gè)線程T1、T2、T3，如何讓他們按順序執(zhí)行？

這是一道面試中常考的并發(fā)編程的代碼題，與它相似的問題有：

三個(gè)線程T1、T2、T3輪流打印ABC，打印n次，如ABCABCABCABC.......
兩個(gè)線程交替打印1-100的奇偶數(shù)
N個(gè)線程循環(huán)打印1-100
......

其實(shí)這類問題本質(zhì)上都是線程通信問題，思路基本上都是一個(gè)線程執(zhí)行完畢，阻塞該線程，喚醒其他線程，按順序執(zhí)行下一個(gè)線程。下面先來看最簡單的，如何按順序執(zhí)行三個(gè)線程。

synchronized+wait/notify

基本思路就是線程A、線程B、線程C三個(gè)線程同時(shí)啟動(dòng)，因?yàn)樽兞?code style="font-size: 14px;overflow-wrap: break-word;padding: 2px 4px;border-radius: 4px;margin-right: 2px;margin-left: 2px;background-color: rgba(27, 31, 35, 0.05);font-family: "Operator Mono", Consolas, Monaco, Menlo, monospace;word-break: break-all;color: rgb(239, 112, 96);">num的初始值為0，所以線程B或線程C拿到鎖后，進(jìn)入while()循環(huán)，然后執(zhí)行wait()方法，線程線程阻塞，釋放鎖。只有線程A拿到鎖后，不進(jìn)入while()循環(huán)，執(zhí)行num++，打印字符A，最后喚醒線程B和線程C。此時(shí)num值為1，只有線程B拿到鎖后，不被阻塞，執(zhí)行num++，打印字符B，最后喚醒線程A和線程C，后面以此類推。

class?Wait_Notify_ACB?{

????private?int?num;
????private?static?final?Object?LOCK?=?new?Object();

????private?void?printABC(int?targetNum)?{
????????????synchronized?(LOCK)?{
????????????????while?(num?%?3?!=?targetNum)?{????//想想這里為什么不能用if代替while，想不起來可以看公眾號(hào)上一篇文章
????????????????????try?{
????????????????????????LOCK.wait();
????????????????????}?catch?(InterruptedException?e)?{
????????????????????????e.printStackTrace();
????????????????????}
????????????????}
????????????????num++;
????????????????System.out.print(Thread.currentThread().getName());
????????????????LOCK.notifyAll();
????????????}
????}
????
????public?static?void?main(String[]?args)?{
????????Wait_Notify_ACB??wait_notify_acb?=?new?Wait_Notify_ACB?();
????????new?Thread(()?->?{
????????????wait_notify_acb.printABC(0);
????????},?"A").start();
????????new?Thread(()?->?{
????????????wait_notify_acb.printABC(1);
????????},?"B").start();
????????new?Thread(()?->?{
????????????wait_notify_acb.printABC(2);
????????},?"C").start();
????}
}

輸入結(jié)果：

ABC
Process?finished?with?exit?code?0

接下來看看第一個(gè)問題，三個(gè)線程T1、T2、T3輪流打印ABC，打印n次。其實(shí)只需要將上述代碼加一個(gè)循環(huán)即可，這里假設(shè)n=10。

class?Wait_Notify_ACB?{

????private?int?num;
????private?static?final?Object?LOCK?=?new?Object();

????private?void?printABC(int?targetNum)?{
????????for?(int?i?=?0;?i?10;?i++)?{
????????????synchronized?(LOCK)?{
????????????????while?(num?%?3?!=?targetNum)?{?//想想這里為什么不能用if代替，想不起來可以看公眾號(hào)上一篇文章
????????????????????try?{
????????????????????????LOCK.wait();
????????????????????}?catch?(InterruptedException?e)?{
????????????????????????e.printStackTrace();
????????????????????}
????????????????}
????????????????num++;
????????????????System.out.print(Thread.currentThread().getName());
????????????????LOCK.notifyAll();
????????????}
????????}
????}
????public?static?void?main(String[]?args)?{
????????Wait_Notify_ACB??wait_notify_acb?=?new?Wait_Notify_ACB?();
????????new?Thread(()?->?{
????????????wait_notify_acb.printABC(0);
????????},?"A").start();
????????new?Thread(()?->?{
????????????wait_notify_acb.printABC(1);
????????},?"B").start();
????????new?Thread(()?->?{
????????????wait_notify_acb.printABC(2);
????????},?"C").start();
????}????
}

輸出結(jié)果：

ABCABCABCABCABCABCABCABCABCABC
Process?finished?with?exit?code?0

下面看第二個(gè)問題，兩個(gè)線程交替打印1-100的奇偶數(shù)，為了減少輸入所占篇幅，這里將100 改成了10。基本思路上面類似，線程odd先拿到鎖——打印數(shù)字——喚醒線程even——阻塞線程odd，以此循環(huán)。

class??Wait_Notify_Odd_Even{

????private?Object?lock?=?new?Object();
????private?volatile?int?count;

????private?void?printOddEven()?{
????????synchronized?(lock)?{
????????????while?(count?10)?{
????????????????try?{
????????????????????System.out.print(?Thread.currentThread().getName()?+?"：");
????????????????????System.out.println(++count);
????????????????????lock.notifyAll();
????????????????????lock.wait();
????????????????}?catch?(InterruptedException?e)?{
????????????????????e.printStackTrace();
????????????????}
????????????}
????????????//防止count=10后，while()循環(huán)不再執(zhí)行，有子線程被阻塞未被喚醒，導(dǎo)致主線程不能退出
????????????lock.notifyAll();
????????}
????}
????
????public?static?void?main(String[]?args)?throws?InterruptedException?{

????????Wait_Notify_Odd_Even?waitNotifyOddEven?=?new?Wait_Notify_Odd_Even();
????????new?Thread(waitNotifyOddEven::printOddEven,?"odd").start();
????????Thread.sleep(10);?//為了保證線程odd先拿到鎖
????????new?Thread(waitNotifyOddEven::printOddEven,?"even").start();
????}
}

運(yùn)行結(jié)果：

odd：1
even：2
odd：3
even：4
odd：5
even：6
odd：7
even：8
odd：9
even：10

再看第三個(gè)問題，N個(gè)線程循環(huán)打印1-100，其實(shí)仔細(xì)想想這個(gè)和三個(gè)線程循環(huán)打印ABC并沒有什么本質(zhì)區(qū)別，只需要加上判斷是否到了打印數(shù)字的最大值的語句即可。假設(shè)N=3，為了能把輸出結(jié)果完全顯示，打印1-10，代碼如下：

class?Wait_Notify_100?{

????private?int?num;
????private?static?final?Object?LOCK?=?new?Object();
????private?int?maxnum?=?10;

????private?void?printABC(int?targetNum)?{
????????while?(true)?{
????????????synchronized?(LOCK)?{
????????????????while?(num?%?3?!=?targetNum)?{?//想想這里為什么不能用if代替，想不起來可以看公眾號(hào)上一篇文章
????????????????????if(num?>=?maxnum){
????????????????????????break;
????????????????????}
????????????????????try?{
????????????????????????LOCK.wait();
????????????????????}?catch?(InterruptedException?e)?{
????????????????????????e.printStackTrace();
????????????????????}
????????????????}
????????????????if(num?>=?maxnum){
????????????????????break;
????????????????}
????????????????num++;
????????????????System.out.println(Thread.currentThread().getName()?+?":?"?+?num);
????????????????LOCK.notifyAll();
????????????}
????????}

????}
????
????public?static?void?main(String[]?args)?{
????????Wait_Notify_100??wait_notify_100?=?new?Wait_Notify_100?();
????????new?Thread(()?->?{
????????????wait_notify_100.printABC(0);
????????},?"thread1").start();
????????new?Thread(()?->?{
????????????wait_notify_100.printABC(1);
????????},?"thread2").start();
????????new?Thread(()?->?{
????????????wait_notify_100.printABC(2);
????????},?"thread3").start();
????}????
}

輸出結(jié)果：

thread1:?1
thread2:?2
thread3:?3
thread1:?4
thread2:?5
thread3:?6
thread1:?7
thread2:?8
thread3:?9
thread1:?10

面試官： 大家都是用的synchronized+wait/notify，你能不能換個(gè)方法解決該問題？

我：好的，我還會(huì)用join()方法

下面介紹的方法只給出第一道題的代碼了，否則太長了，相信大家可以舉一反三

join()

join()方法：在A線程中調(diào)用了B線程的join()方法時(shí)，表示只有當(dāng)B線程執(zhí)行完畢時(shí)，A線程才能繼續(xù)執(zhí)行。基于這個(gè)原理，我們使得三個(gè)線程按順序執(zhí)行，然后循環(huán)多次即可。無論線程1、線程2、線程3哪個(gè)先執(zhí)行，最后執(zhí)行的順序都是線程1——>線程2——>線程3。代碼如下：

class?Join_ABC?{

????public?static?void?main(String[]?args)?throws?InterruptedException?{
????????for?(int?i?=?0;?i?10;?i++)?{
????????????Thread?t1?=?new?Thread(new?printABC(null),"A");
????????????Thread?t2?=?new?Thread(new?printABC(t1),"B");
????????????Thread?t3?=?new?Thread(new?printABC(t2),"C");
????????????t0.start();
????????????t1.start();
????????????t2.start();
????????????Thread.sleep(10);?//這里是要保證只有t1、t2、t3為一組，進(jìn)行執(zhí)行才能保證t1->t2->t3的執(zhí)行順序。
????????}

????}

????static?class?printABC?implements?Runnable{
????????private?Thread?beforeThread;
????????public?printABC(Thread?beforeThread)?{
????????????this.beforeThread?=?beforeThread;
????????}
????????@Override
????????public?void?run()?{
????????????if(beforeThread!=null)?{
????????????????try?{
????????????????????beforeThread.join();?
????????????????????System.out.print(Thread.currentThread().getName());
????????????????}catch(Exception?e){
????????????????????e.printStackTrace();
????????????????}
????????????}else?{
????????????????System.out.print(Thread.currentThread().getName());
????????????}

????????}
????}
}

輸出結(jié)果：

ABCABCABCABCABCABCABCABCABCABC

面試官： 還會(huì)其他方法嗎？

我：還會(huì)使用Lock解決該問題。

Lock

該方法很容易理解，不管哪個(gè)線程拿到鎖，只有符合條件的才能打印。代碼如下：

?class?Lock_ABC?{

????private?int?num;???//?當(dāng)前狀態(tài)值：保證三個(gè)線程之間交替打印
????private?Lock?lock?=?new?ReentrantLock();


????private?void?printABC(int?targetNum)?{
????????for?(int?i?=?0;?i?10;?)?{
????????????lock.lock();
????????????if?(num?%?3?==?targetNum)?{
????????????????num++;
????????????????i++;
????????????????System.out.print(Thread.currentThread().getName());
????????????}
????????????lock.unlock();
????????}
????}

????public?static?void?main(String[]?args)?{
????????Lock_ABC?lockABC?=?new?Lock_ABC();

????????new?Thread(()?->?{
????????????lockABC.printABC(0);
????????},?"A").start();

????????new?Thread(()?->?{
????????????lockABC.printABC(1);
????????},?"B").start();

????????new?Thread(()?->?{
????????????lockABC.printABC(2);
????????},?"C").start();
????}
}

輸出結(jié)果：

ABCABCABCABCABCABCABCABCABCABC

面試官： 該方法存在什么問題，可以進(jìn)一步優(yōu)化嗎

我：可以使用Lock+Condition實(shí)現(xiàn)對(duì)線程的精準(zhǔn)喚醒，減少對(duì)同步鎖的無意義競爭，浪費(fèi)資源。

Lock+Condition

該思路和synchronized+wait/notify方法的很像，synchronized對(duì)應(yīng)lock，await/signal方法對(duì)應(yīng)wait/notify方法。下面的代碼為了能精準(zhǔn)地喚醒下一個(gè)線程，創(chuàng)建了多個(gè)Condition對(duì)象。

class?LockConditionABC?{

????private?int?num;
????private?static?Lock?lock?=?new?ReentrantLock();
????private?static?Condition?c1?=?lock.newCondition();
????private?static?Condition?c2?=?lock.newCondition();
????private?static?Condition?c3?=?lock.newCondition();

????private?void?printABC(int?targetNum,?Condition?currentThread,?Condition?nextThread)?{
????????for?(int?i?=?0;?i?10;?)?{
????????????lock.lock();
????????????try?{
????????????????while?(num?%?3?!=?targetNum)?{
????????????????????currentThread.await();??//阻塞當(dāng)前線程
????????????????}
????????????????num++;
????????????????i++;
????????????????System.out.print(Thread.currentThread().getName());
????????????????nextThread.signal();????//喚醒下一個(gè)線程，而不是喚醒所有線程
????????????}?catch?(Exception?e)?{
????????????????e.printStackTrace();
????????????}?finally?{
????????????????lock.unlock();
????????????}
????????}
????}

????public?static?void?main(String[]?args)?{
????????LockConditionABC?print?=?new?LockConditionABC();
????????new?Thread(()?->?{
????????????print.printABC(0,?c1,?c2);
????????},?"A").start();
????????new?Thread(()?->?{
????????????print.printABC(1,?c2,?c3);
????????},?"B").start();
????????new?Thread(()?->?{
????????????print.printABC(2,?c3,?c1);
????????},?"C").start();
????}
}

輸出結(jié)果：

ABCABCABCABCABCABCABCABCABCABC

面試官： 除了該方法，還有什么方法可以避免喚醒其他無意義的線程避免資源浪費(fèi)？

我：可以通過使用信號(hào)量來實(shí)現(xiàn)。

Semaphore

Semaphore：用來控制同時(shí)訪問某個(gè)特定資源的操作數(shù)量，或者同時(shí)執(zhí)行某個(gè)指定操作的數(shù)量。Semaphore內(nèi)部維護(hù)了一個(gè)計(jì)數(shù)器，其值為可以訪問的共享資源的個(gè)數(shù)。

一個(gè)線程要訪問共享資源，先使用acquire()方法獲得信號(hào)量，如果信號(hào)量的計(jì)數(shù)器值大于等于1，意味著有共享資源可以訪問，則使其計(jì)數(shù)器值減去1，再訪問共享資源。如果計(jì)數(shù)器值為0,線程進(jìn)入休眠。

當(dāng)某個(gè)線程使用完共享資源后，使用release()釋放信號(hào)量，并將信號(hào)量內(nèi)部的計(jì)數(shù)器加1，之前進(jìn)入休眠的線程將被喚醒并再次試圖獲得信號(hào)量。

代碼如下：

class?SemaphoreABC?{

????private?static?Semaphore?s1?=?new?Semaphore(1);?//因?yàn)橄葓?zhí)行線程A，所以這里設(shè)s1的計(jì)數(shù)器為1
????private?static?Semaphore?s2?=?new?Semaphore(0);
????private?static?Semaphore?s3?=?new?Semaphore(0);


????private?void?printABC(Semaphore?currentThread,?Semaphore?nextThread)?{
????????for?(int?i?=?0;?i?10;?i++)?{
????????????try?{
????????????????currentThread.acquire();???????//阻塞當(dāng)前線程，即信號(hào)量的計(jì)數(shù)器減1為0
????????????????System.out.print(Thread.currentThread().getName());
????????????????nextThread.release();??????????//喚醒下一個(gè)線程，即信號(hào)量的計(jì)數(shù)器加1

????????????}?catch?(InterruptedException?e)?{
????????????????e.printStackTrace();
????????????}
????????}
????}

????public?static?void?main(String[]?args)?throws?InterruptedException?{
????????SemaphoreABC?printer?=?new?SemaphoreABC();
????????new?Thread(()?->?{
????????????printer.printABC(s1,?s2);
????????},?"A").start();
????????Thread.sleep(10);
????????new?Thread(()?->?{
????????????printer.printABC(s2,?s3);
????????},?"B").start();
????????Thread.sleep(10);
????????new?Thread(()?->?{
????????????printer.printABC(s3,?s1);
????????},?"C").start();
????}
}

輸出結(jié)果：

ABCABCABCABCABCABCABCABCABCABC

面試官： 除了上述五種方法，還有其他方法解決多線程交替打印問題嗎

我：還有LockSupport、CountDownLatch、AtomicInteger等等。

面試官： 那如何實(shí)現(xiàn)三個(gè)線程循環(huán)打印ACB，其中A打印兩次，B打印三次，C打印四次呢？

我：......

面試官：如何用兩個(gè)線程交叉打印數(shù)字和字符呢？例如A1B2C3......Z26

我：......

真實(shí)的面試中，面試官肯定不會(huì)讓用這么多方法解決多線程交替打印問題，大家記住一兩種就可以了，上面中的面試場景純屬瞎編。大家可以思考下后面兩個(gè)升級(jí)版的問題，原理都是相通的。

面試官：請(qǐng)用五種方法來實(shí)現(xiàn)多線程交替打印問題

三個(gè)線程T1、T2、T3，如何讓他們按順序執(zhí)行？

synchronized+wait/notify

join()

Lock

Lock+Condition

Semaphore