最近
发现公司有份笔试试卷中有道多
线程的
题目:
有三个线程分别打印A、B、C,请用多线程编程实现,在屏幕上循环打印10次ABCABC…
这个最早好像是迅雷的面试题目吧,看到了然后就想重温一下这个题目的
解决方法。
在本文中,给出了五种这个题目的解决方法:
- 使用sleep
- 使用synchronized, wait和notifyAll
- 使用Lock 和 Condition
- 使用Semaphore
- 使用AtomicInteger
下面依次给出每种解决方案的代码:
使用sleep
class="java">package my.thread.test;
/**
* @author Eric
*/
public class SleepExample extends Thread {
private static int currentCount = 0;
public SleepExample(String name) {
this.setName(name);
}
@Override
public void run() {
while (currentCount < 30) {
switch (currentCount % 3) {
case 0:
if ("A".equals(getName())) {
printAndIncrease();
}
break;
case 1:
if ("B".equals(getName())) {
printAndIncrease();
}
break;
case 2:
if ("C".equals(getName())) {
printAndIncrease();
}
break;
}
}
}
private void printAndIncrease() {
print();
increase();
}
private void print() {
System.out.println(getName());
if ("C".equals(getName())) {
System.out.println();
}
}
private void increase() {
currentCount++;
}
public static void main(String[] args) {
new SleepExample("A").start();
new SleepExample("B").start();
new SleepExample("C").start();
}
}
使用synchronized, wait和notifyAll
package my.thread.test;
import java.util.concurrent.ExecutorService;
import java.util.concurrent.Executors;
public class PrintThreadExample {
public static void main(String[] args) {
PrintThreadExample example = new PrintThreadExample();
LetterPrinter letterPrinter = example.new LetterPrinter();
ExecutorService service = Executors.newFixedThreadPool(3);
service.execute(example.new PrintRunnable(letterPrinter, 'A'));
service.execute(example.new PrintRunnable(letterPrinter, 'B'));
service.execute(example.new PrintRunnable(letterPrinter, 'C'));
service.shutdown();
}
private class LetterPrinter {
private char letter = 'A';
public void print() {
System.out.println(letter);
if ('C' == letter) {
System.out.println();
}
}
public void nextLetter() {
switch (letter) {
case 'A':
letter = 'B';
break;
case 'B':
letter = 'C';
break;
case 'C':
letter = 'A';
break;
}
}
/**
* @return the letter
*/
public char getLetter() {
return letter;
}
}
private class PrintRunnable implements Runnable {
private LetterPrinter letterPrinter = null;
private char letter = ' ';
/**
* @param letterPrinter
* @param letter
*/
public PrintRunnable(LetterPrinter letterPrinter, char letter) {
super();
this.letterPrinter = letterPrinter;
this.letter = letter;
}
public void run() {
for (int i = 0; i < 10; i++) {
synchronized (letterPrinter) {
while (letter != letterPrinter.getLetter()) {
try {
letterPrinter.wait();
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
letterPrinter.print();
letterPrinter.nextLetter();
letterPrinter.notifyAll();
}
}
}
}
}
JDK 1.5 引入J.U.C包之后,也给我们提供了更多实现多线程程序的选择: Condition, 原子类AtomicInteger以及Semaphore等。
使用Lock 和 Condition
package my.thread.test;
import java.util.concurrent.ExecutorService;
import java.util.concurrent.Executors;
import java.util.concurrent.locks.Condition;
import java.util.concurrent.locks.Lock;
import java.util.concurrent.locks.ReentrantLock;
import java.util.logging.Logger;
/**
* 题目:有三个线程分别打印A、B、C,请用多线程编程实现,在屏幕上循环打印10次ABCABC…
*
* 本程序采用Lock和Condition来实现。
*
* @author Eric
*
*/
public class ConditionExample {
private Lock lock = new ReentrantLock();
private Condition conditionA = lock.newCondition();
private Condition conditionB = lock.newCondition();
private Condition conditionC = lock.newCondition();
/** 当前线程的名字 */
private char currentThreadName = 'A';
private static final Logger logger = Logger
.getLogger("my.thread.test.OrderPrintTest");
public static void main(String[] args) {
ConditionExample ce = new ConditionExample();
ExecutorService service = Executors.newFixedThreadPool(3);
service.execute(ce.new ThreadA());
service.execute(ce.new ThreadB());
service.execute(ce.new ThreadC());
service.shutdown();
}
private class ThreadA implements Runnable {
public void run() {
for (int i = 0; i < 10; i++) {
lock.lock();
try {
while (currentThreadName != 'A') {
try {
/*
* 如果当前线程名字不是A,那么ThreadA就处理等待状态
*/
conditionA.await();
} catch (InterruptedException e) {
logger.severe(e.getLocalizedMessage());
}
}
/*
* 打印出第几遍以及A信息
*/
System.out.println(String.format("第%d遍", i + 1));
System.out.println("A");
/*
* 将当前线程名置为B, 然后通知ThreadB执行
*/
currentThreadName = 'B';
conditionB.signal();
} finally {
lock.unlock();
}
}
}
}
private class ThreadB implements Runnable {
public void run() {
for (int i = 0; i < 10; i++) {
lock.lock();
try {
while (currentThreadName != 'B') {
try {
/*
* 如果当前线程名字不是B,那么ThreadB就处理等待状态
*/
conditionB.await();
} catch (InterruptedException e) {
logger.severe(e.getLocalizedMessage());
}
}
/*
* 打印信息B
*/
System.out.println("B");
/*
* 将当前线程值置为C 并通过ThreadC来执行
*/
currentThreadName = 'C';
conditionC.signal();
} finally {
lock.unlock();
}
}
}
}
private class ThreadC implements Runnable {
public void run() {
for (int i = 0; i < 10; i++) {
lock.lock();
try {
while (currentThreadName != 'C') {
try {
/*
* 如果当前线程名字不是C,那么ThreadC就处理等待状态
*/
conditionC.await();
} catch (InterruptedException e) {
logger.severe(e.getLocalizedMessage());
}
}
/*
* 打印信息C
*/
System.out.println("C");
System.out.println();
/*
* 将当前线程值置为A 并通过ThreadA来执行
*/
currentThreadName = 'A';
conditionA.signal();
} finally {
lock.unlock();
}
}
}
}
}
使用Semaphore
package my.thread.test;
import java.util.concurrent.ExecutorService;
import java.util.concurrent.Executors;
import java.util.concurrent.Semaphore;
public class SemaphoresExample {
private Semaphore semaphoresA = new Semaphore(1);
private Semaphore semaphoresB = new Semaphore(0);
private Semaphore semaphoresC = new Semaphore(0);
public static void main(String[] args) {
SemaphoresExample example = new SemaphoresExample();
ExecutorService service = Executors.newFixedThreadPool(3);
service.execute(example.new RunnableA());
service.execute(example.new RunnableB());
service.execute(example.new RunnableC());
service.shutdown();
}
private class RunnableA implements Runnable {
public void run() {
for (int i = 0; i < 10; i++) {
try {
semaphoresA.acquire();
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
System.out.println(String.format("第%d遍", i + 1));
System.out.println("A");
semaphoresB.release();
}
}
}
private class RunnableB implements Runnable {
public void run() {
for (int i = 0; i < 10; i++) {
try {
semaphoresB.acquire();
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
System.out.println("B");
semaphoresC.release();
}
}
}
private class RunnableC implements Runnable {
public void run() {
for (int i = 0; i < 10; i++) {
try {
semaphoresC.acquire();
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
System.out.println("C");
System.out.println();
semaphoresA.release();
}
}
}
}
使用AtomicInteger
package my.thread.test;
import java.util.concurrent.ExecutorService;
import java.util.concurrent.Executors;
import java.util.concurrent.atomic.AtomicInteger;
public class AtomicIntegerExample {
private AtomicInteger sycValue = new AtomicInteger(0);
private static final int MAX_SYC_VALUE = 3 * 10;
public static void main(String[] args) {
AtomicIntegerExample example = new AtomicIntegerExample();
ExecutorService service = Executors.newFixedThreadPool(3);
service.execute(example.new RunnableA());
service.execute(example.new RunnableB());
service.execute(example.new RunnableC());
service.shutdown();
}
private class RunnableA implements Runnable {
public void run() {
while (sycValue.get() < MAX_SYC_VALUE) {
if (sycValue.get() % 3 == 0) {
System.out.println(String.format("第%d遍",
sycValue.get() / 3 + 1));
System.out.println("A");
sycValue.getAndIncrement();
}
}
}
}
private class RunnableB implements Runnable {
public void run() {
while (sycValue.get() < MAX_SYC_VALUE) {
if (sycValue.get() % 3 == 1) {
System.out.println("B");
sycValue.getAndIncrement();
}
}
}
}
private class RunnableC implements Runnable {
public void run() {
while (sycValue.get() < MAX_SYC_VALUE) {
if (sycValue.get() % 3 == 2) {
System.out.println("C");
System.out.println();
sycValue.getAndIncrement();
}
}
}
}
}
转载请注明出处http://mouselearnjava.iteye.com/blog/1949228