編寫一個程序，開啓3個線程，這三個線程的ID分別爲A、B、C，每個線程將自己的ID在屏幕上打印10遍，要求輸出的結果必須按順序顯示。

編寫一個程序，開啓3個線程，這三個線程的ID分別爲A、B、C，每個線程將自己的ID在屏幕上打印10遍，要求輸出的結果必須按順序顯示。

一道Java併發相關的面試題。本文使用Lock+Condition實現。

package com.xudc.juc;

import java.util.concurrent.locks.Condition;
import java.util.concurrent.locks.Lock;
import java.util.concurrent.locks.ReentrantLock;

/**
 * 編寫一個程序，開啓3個線程，這三個線程的ID分別爲A、B、C，每個線程將自己的ID在屏幕上打印10遍，要求輸出的結果必須按順序顯示。
 * 如：ABCABCABCABC...依次遞歸
 * @author xudc
 */
public class TestAlternate {
    public static void main(String[] args) {
        Alternate alternate = new Alternate();
        final int count = 10;
        new Thread(() -> {
            for (int i = 0; i < count; i++) {
                alternate.printA();
            }
        },"A").start();
        new Thread(() -> {
            for (int i = 0; i < count; i++) {
                alternate.printB();
            }
        },"B").start();
        new Thread(() -> {
            for (int i = 0; i < count; i++) {
                alternate.printC();
            }
        },"C").start();
    }
}

class Alternate{
    private final Lock lock = new ReentrantLock();
    private Condition conditionA = lock.newCondition();
    private Condition conditionB = lock.newCondition();
    private Condition conditionC = lock.newCondition();

    private int number = 1;

    public void printA(){
        lock.lock();
        try {
            while (number != 1) {
                conditionA.await();
            }
            System.out.print("A");
            number = 2;
            conditionB.signal();
        } catch (InterruptedException e) {
            e.printStackTrace();
        } finally {
            lock.unlock();
        }
    }

    public void printB(){
        lock.lock();
        try {
            while (number != 2) {
                conditionB.await();
            }
            System.out.print("B");
            number = 3;
            conditionC.signal();
        } catch (InterruptedException e) {
            e.printStackTrace();
        } finally {
            lock.unlock();
        }
    }

    public void printC(){
        lock.lock();
        try {
            while (number != 3) {
                conditionC.await();
            }
            System.out.print("C");
            number = 1;
            conditionA.signal();
        } catch (InterruptedException e) {
            e.printStackTrace();
        } finally {
            lock.unlock();
        }
    }
}

運行結果：

ABCABCABCABCABCABCABCABCABCABC

原文地址：https://xudc.tech/2019/06/01/thread-interview-1/