大厂面试题：两个线程交叉打印，一个打印数字，一个打印小写字母，结果为 1A2B3C4D5E6F

我不是女神ヾ 2023-07-15 08:45 9阅读 0赞

前言：这道题主要考察的就是，多线程之间的通信、唤醒、切换问题。

B站上看到马士兵老师的视频：[这道大厂面试题区分平庸与卓越][Link 1]

部分解法整理如下：

--------------------

### LockSupport实现方式 ###

/**
     * @ClassName: LockSupport
     * @description: 两个线程交叉打印，一个打印数字，一个打印小写字母，结果为 1A2B3C4D5E6F
     * LockSupport实现
     * @author: XZQ
     * @create: 2020/3/15 15:24
     **/
    public class LockSupportDemo {
    
        static Thread t1 = null, t2 = null;
    
        public static void main(String[] args) {
            char[] aI = "1234567".toCharArray();
            char[] aC = "ABCDEFG".toCharArray();
    
            t1 = new Thread(() -> {
                for (char c : aI) {
                    System.out.print(c);
                    LockSupport.unpark(t2);//叫醒t2 叫醒指定线程
                    LockSupport.park();//T1阻塞 当前线程阻塞
                }
            }, "t1");
    
    
            t2 = new Thread(() -> {
                for (char c : aC) {
                    LockSupport.park();//t2阻塞
                    System.out.print(c);
                    LockSupport.unpark(t1);//叫醒t1
                }
            }, "t2");
    
            t1.start();
            t2.start();
        }
    }

### CAS实现方式 ###

/**
     * @ClassName: CASDemo
     * @description: 两个线程交叉打印，一个打印数字，一个打印小写字母，结果为 1A2B3C4D5E6F
     * CAS实现
     * @author: XZQ
     * @create: 2020/3/15 15:56
     **/
    public class CASDemo {
        enum ReadToRun {T1, T2}//枚举 规范参数的形式
    
        static volatile ReadToRun r = ReadToRun.T1;//保证可见性
    
        public static void main(String[] args) {
            char[] aI = "1234567".toCharArray();
            char[] aC = "ABCDEFG".toCharArray();
    
            new Thread(() -> {
                for (char c : aI) {
                    while (r != ReadToRun.T1) {
                    }
                    System.out.print(c);
                    r = ReadToRun.T2;
                }
            }, "t1").start();
    
            new Thread(() -> {
                for (char c : aC) {
                    while (r != ReadToRun.T2) {
                    }
                    System.out.print(c);
                    r = ReadToRun.T1;
                }
            }, "t2").start();
        }
    }

### ReentrantLock\_Condition实现方式 ###

/**
     * @ClassName: ReentrantLockDemo
     * @description: 两个线程交叉打印，一个打印数字，一个打印小写字母，结果为 1A2B3C4D5E6F
     * ReentrantLock&Condition 实现方式
     * 优点:支持指定线程唤醒和指定线程等待
     * @author: XZQ
     * @create: 2020/3/15 16:27
     **/
    public class ReentrantLock_ConditionDemo {
    
        public static void main(String[] args) {
            char[] aI = "1234567".toCharArray();
            char[] aC = "ABCDEFG".toCharArray();
    
            Lock lock = new ReentrantLock();//默认非公平锁
            Condition condition1 = lock.newCondition();//队列
            Condition condition2 = lock.newCondition();
    
            new Thread(() -> {
                try {
                    lock.lock();
                    for (char c : aI) {
                        System.out.print(c);
                        condition2.signal();
                        condition1.await();
                    }
                    condition2.signal();//必须 否则无法停止程序
                } catch (Exception e) {
                    e.printStackTrace();
                } finally {
                    lock.unlock();
                }
            }, "t1").start();
    
            new Thread(() -> {
                try {
                    lock.lock();
                    for (char c : aC) {
                        System.out.print(c);
                        condition1.signal();
                        condition2.await();
                    }
                    condition1.signal();//必须 否则无法停止程序
                } catch (Exception e) {
                    e.printStackTrace();
                } finally {
                    lock.unlock();
                }
            }, "t2").start();
        }
    }

### ReentrantLock实现方式 ###

/**
     * @ClassName: ReentrantLockDemo
     * @description: 两个线程交叉打印，一个打印数字，一个打印小写字母，结果为 1A2B3C4D5E6F
     * ReentrantLock 实现方式
     * @author: XZQ
     * @create: 2020/3/15 16:27
     **/
    public class ReentrantLockDemo {
    
        public static void main(String[] args) {
            char[] aI = "1234567".toCharArray();
            char[] aC = "ABCDEFG".toCharArray();
    
            ReentrantLock reentrantLock = new ReentrantLock();//默认非公平锁
            Condition condition = reentrantLock.newCondition();
    
            new Thread(() -> {
                try {
                    reentrantLock.lock();
                    for (char c : aI) {
                        System.out.print(c);
                        condition.signal();
                        condition.await();
                    }
                    condition.signal();//必须 否则无法停止程序
                } catch (Exception e) {
                    e.printStackTrace();
                } finally {
                    reentrantLock.unlock();
                }
            }, "t1").start();
    
            new Thread(() -> {
                try {
                    reentrantLock.lock();
                    for (char c : aC) {
                        System.out.print(c);
                        condition.signal();
                        condition.await();
                    }
                    condition.signal();//必须 否则无法停止程序
                } catch (Exception e) {
                    e.printStackTrace();
                } finally {
                    reentrantLock.unlock();
                }
            }, "t2").start();
        }
    }

### Synchronized实现方式 ###

/**
     * @ClassName: SynchronizedDemo
     * @description: 两个线程交叉打印，一个打印数字，一个打印小写字母，结果为 1A2B3C4D5E6F
     * Synchronized实现
     * @author: XZQ
     * @create: 2020/3/15 16:15
     **/
    public class SynchronizedDemo {
        static Object object = new Object();
    
        public static void main(String[] args) {
            char[] aI = "1234567".toCharArray();
            char[] aC = "ABCDEFG".toCharArray();
    
            new Thread(() -> {
                synchronized (object) {
                    for (char c : aI) {
                        System.out.print(c);
                        try {
                            object.notify();//先notify再wait
                            object.wait();
                        } catch (InterruptedException e) {
                            e.printStackTrace();
                        }
                    }
                    object.notify();//必须 否则无法停止程序
                }
            }, "t1").start();
    
            new Thread(() -> {
                synchronized (object) {
                    for (char c : aC) {
                        System.out.print(c);
                        try {
                            object.notify();
                            object.wait();
                        } catch (InterruptedException e) {
                            e.printStackTrace();
                        }
                    }
                    object.notify();//必须 否则无法停止程序
                }
            }, "t2").start();
        }
    }

### TransferQueue 实现方式 ###

/**
     * @ClassName: TransferQueueDemo
     * @description: 两个线程交叉打印，一个打印数字，一个打印小写字母，结果为 1A2B3C4D5E6F
     * TransferQueue 实现方式
     * @author: XZQ
     * @create: 2020/3/15 16:54
     **/
    public class TransferQueueDemo {
        public static void main(String[] args) {
            TransferQueue<Character> queue = new LinkedTransferQueue<>();
            char[] aI = "1234567".toCharArray();
            char[] aC = "ABCDEFG".toCharArray();
    
            new Thread(() -> {
                try {
                    for (char c : aC) {
                        System.out.print(queue.take());
                        queue.transfer(c);
                    }
                } catch (Exception e) {
                    e.printStackTrace();
                }
    
            }, "t1").start();
    
            new Thread(() -> {
                try {
                    for (char c : aI) {
                        queue.transfer(c);
                        System.out.print(queue.take());
                    }
                } catch (Exception e) {
                    e.printStackTrace();
                }
    
            }, "t2").start();
        }
    }

除此之外还有阻塞队列等等一系列解法。

[Link 1]: https://www.bilibili.com/video/av92399114