有以下需要：
两个线程，须要打印字母和数字，格局A1B2C3 …
这个问题波及到线程的期待，唤醒，线程间通信等常识。
上面看看实现代码：
办法1 LockSupport
import java.util.concurrent.locks.LockSupport;/** * @author liming * @date 2020/10 * @description 交替打印 A1B2C3 ... */public class AlternatePrint { static Thread t1 = null, t2 = null; public static void main(String[] args) { char[] aI = "1234567".toCharArray(); char[] aC = "ABCDEFG".toCharArray(); t1 = new Thread(new Runnable() { @Override public void run() { for (int i = 0; i < aC.length; i++) { // 起始先打印一个字母 System.out.println(aC[i]); // 打印完唤醒t2打印数字 LockSupport.unpark(t2); // 本人阻塞，期待唤醒 LockSupport.park(); } } }); t2 = new Thread(new Runnable() { @Override public void run() { for (int i = 0; i < aI.length; i++) { // 起始先阻塞期待 LockSupport.park(); // 被唤醒后打印数字 System.out.println(aI[i]); // 唤醒t1 LockSupport.unpark(t1); } } }); t1.start(); t2.start(); }}办法2 synchronized
import org.junit.Test;/** * @author liming * @date 2020/10/14 * @description 交替打印 A1B2C3 ... */public class AlternatePrint { static Thread t1 = null, t2 = null; /** * 应用 synchronized */ @Test public void alternatePrint() { Object lock = new Object(); char[] aI = "1234567".toCharArray(); char[] aC = "ABCDEFG".toCharArray(); t1 = new Thread(new Runnable() { @Override public void run() { for (int i = 0; i < aC.length; i++) { synchronized (lock) { System.out.println(aC[i]); lock.notify(); try { lock.wait(); } catch (InterruptedException e) { e.printStackTrace(); } } } } }); t2 = new Thread(new Runnable() { @Override public void run() { for (int i = 0; i < aI.length; i++) { synchronized (lock) { System.out.println(aI[i]); lock.notify(); try { lock.wait(); } catch (InterruptedException e) { e.printStackTrace(); } } } } }); t1.start(); t2.start(); }}测试后果：