Java线程（练习题）

Exercise

创建三个线程：一个线程打印 100个A,一个线程打印 100 个 B ，一个线程打印 100个C 输出效果：ABC ABC ABC…交替打印

package com.kane.exercise01;public class PrintABC implements Runnable {private static final Object lock = new Object();private static int state = 0;private String str;private int num;public PrintABC(String str, int num) {this.str = str;this.num = num;}@Overridepublic void run() {for (int i = 0; i < 100; i++) {synchronized (lock) {while (state % 3 != num) {try {lock.wait();} catch (InterruptedException e) {e.printStackTrace();}}System.out.print(str);state++;lock.notifyAll(); // 唤醒其他线程}if(state%3 == 0) {System.out.println("");}}}
}

package com.kane.exercise01;public class Demo1 {public static void main(String[] args) {Thread t1 = new Thread(new PrintABC("A",0));Thread t2 = new Thread(new PrintABC("B",1));Thread t3 = new Thread(new PrintABC("C",2));t1.start();t2.start();t3.start();}
}

编写两个线程，一个 线程打印1-52，另一个线程打印字母A-Z打印顺序是12A34B…,即按照整数和字母的顺序从小到大打印，并且每打印两个整数后，打印一个字母，交替循环打印

package com.kane.exercise02;import java.util.concurrent.Semaphore;public class Demo {private static final Semaphore Numbersemaphore = new Semaphore(1);private static final Semaphore Dansemaphore = new Semaphore(0);public static void main(String[] args) {Thread t1 = new Thread(new NumberPrint());Thread t2 = new Thread(new DanPrint());t1.start();t2.start();}static class NumberPrint implements Runnable {@Overridepublic void run() {for (int i = 1; i < 53; i++) {try {Numbersemaphore.acquire();System.out.print(i + " ");if (i%2== 0) {Dansemaphore.release();} else {Numbersemaphore.release();}} catch (InterruptedException e) {throw new RuntimeException(e);}}}}static class DanPrint implements Runnable {@Overridepublic void run() {for (char c = 'A'; c <= 'Z'; c++) {try {Dansemaphore.acquire();System.out.print(c);System.out.println("");Numbersemaphore.release();} catch (InterruptedException e) {Thread.currentThread().interrupt();}}}}}