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原题
package com.fei;
/**
* 现有的程序代码模拟产生了16个日志对象,并且需要运行16秒才能打印完这些日志,
* 请在程序中增加4个线程去调用parseLog()方法来分头打印这16个日志对象,
* 程序只需要运行4秒即可打印完这些日志对象。
*
*/
public class Test {
public static void main(String[] args) {
System.out.println("begin:" + (System.currentTimeMillis() / 1000));
/*
* 模拟处理16行日志,下面的代码产生了16个日志对象,当前代码需要运行16秒才能打印完这些日志。
* 修改程序代码,开四个线程让这16个对象在4秒钟打完。
*/
for (int i = 0; i < 16; i++) { // 这行代码不能改动
final String log = "" + (i + 1);// 这行代码不能改动
{
Test.parseLog(log);
}
}
}
// parseLog方法内部的代码不能改动
public static void parseLog(String log) {
System.out.println(log + ":" + (System.currentTimeMillis() / 1000));
try {
Thread.sleep(1000);
} catch (InterruptedException e) {
e.printStackTrace();
}
}
}
运行结果
begin:1374236125
1:1374236125
2:1374236126
3:1374236127
4:1374236129
5:1374236130
6:1374236131
7:1374236132
8:1374236133
9:1374236134
10:1374236135
11:1374236136
12:1374236137
13:1374236138
14:1374236139
15:1374236140
16:1374236141
修改后:
package com.fei;
import java.util.concurrent.ArrayBlockingQueue;
import java.util.concurrent.BlockingQueue;
import java.util.concurrent.SynchronousQueue;
public class Test {
public static void main(String[] args){
// final SynchronousQueue<String> queue = new SynchronousQueue<String>();
final BlockingQueue<String> queue = new ArrayBlockingQueue<String>(1);
for(int i=0;i<4;i++){
new Thread(new Runnable(){
@Override
public void run() {
while(true){
try {
String log = queue.take();
parseLog(log);
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
}
}).start();
}
System.out.println("begin:"+(System.currentTimeMillis()/1000));
/*模拟处理16行日志,下面的代码产生了16个日志对象,当前代码需要运行16秒才能打印完这些日志。
修改程序代码,开四个线程让这16个对象在4秒钟打完。
*/
for(int i=0;i<16;i++){ //这行代码不能改动
final String log = ""+(i+1);//这行代码不能改动
{
try {
queue.put(log);
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
//Test.parseLog(log);
}
}
}
//parseLog方法内部的代码不能改动
public static void parseLog(String log){
System.out.println(log+":"+(System.currentTimeMillis()/1000));
try {
Thread.sleep(1000);
} catch (InterruptedException e) {
e.printStackTrace();
}
}
}
运行结果
begin:1374236852
1:1374236852
2:1374236852
4:1374236852
3:1374236852
5:1374236853
6:1374236853
7:1374236853
8:1374236853
9:1374236854
10:1374236854
11:1374236854
12:1374236854
13:1374236855
14:1374236855
15:1374236855
16:1374236855
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题目二:
原题
package com.fei;
/**
* 现成程序中的Test类中的代码在不断地产生数据,然后交给TestDo.doSome()方法去处理,
* 就好像生产者在不断地产生数据,消费者在不断消费数据。
* 请将程序改造成有10个线程来消费生成者产生的数据,
* 这些消费者都调用TestDo.doSome()方法去进行处理,
* 故每个消费者都需要一秒才能处理完,
* 程序应保证这些消费者线程依次有序地消费数据,
* 只有上一个消费者消费完后,下一个消费者才能消费数据,
* 下一个消费者是谁都可以,但要保证这些消费者线程拿到的数据是有顺序的。
*
*/
public class Test {
public static void main(String[] args) {
System.out.println("begin:" + (System.currentTimeMillis() / 1000));
for (int i = 0; i < 10; i++) { // 这行不能改动
String input = i + ""; // 这行不能改动
String output = TestDo.doSome(input);
System.out.println(Thread.currentThread().getName() + ":" + output);
}
}
}
// 不能改动此TestDo类
class TestDo {
public static String doSome(String input) {
try {
Thread.sleep(1000);
} catch (InterruptedException e) {
e.printStackTrace();
}
String output = input + ":" + (System.currentTimeMillis() / 1000);
return output;
}
}
运行结果
begin:1374237199
main:0:1374237200
main:1:1374237201
main:2:1374237202
main:3:1374237203
main:4:1374237204
main:5:1374237205
main:6:1374237206
main:7:1374237207
main:8:1374237208
main:9:1374237209
修改:
package com.fei;
import java.util.concurrent.ArrayBlockingQueue;
import java.util.concurrent.BlockingQueue;
import java.util.concurrent.Semaphore;
import java.util.concurrent.SynchronousQueue;
public class Test02 {
public static void main(String[] args) {
final Semaphore semaphore = new Semaphore(1);
// final BlockingQueue<String> queue =new ArrayBlockingQueue<String>(1);
final SynchronousQueue<String> queue = new SynchronousQueue<String>();
for (int i = 0; i < 10; i++) {
new Thread(new Runnable() {
@Override
public void run() {
try {
semaphore.acquire();
String input = queue.take();
String output = TestDo.doSome(input);
System.out.println(Thread.currentThread().getName()
+ ":" + output);
semaphore.release();
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
}).start();
}
System.out.println("begin:" + (System.currentTimeMillis() / 1000));
for (int i = 0; i < 10; i++) { // 这行不能改动
String input = i + ""; // 这行不能改动
try {
queue.put(input);
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
}
}
// 不能改动此TestDo类
class TestDo {
public static String doSome(String input) {
try {
Thread.sleep(1000);
} catch (InterruptedException e) {
e.printStackTrace();
}
String output = input + ":" + (System.currentTimeMillis() / 1000);
return output;
}
}
运行结果
begin:1374237419
Thread-0:0:1374237420
Thread-2:1:1374237421
Thread-4:2:1374237422
Thread-6:3:1374237423
Thread-8:4:1374237424
Thread-1:5:1374237425
Thread-3:6:1374237426
Thread-5:7:1374237427
Thread-7:8:1374237428
Thread-9:9:1374237429
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题目三:
原题:
package syn;
/**
* 现有程序同时启动了4个线程去调用TestDo.doSome(key, value)方法,由于TestDo.doSome(key, value)方法内的代码是先暂停1秒,
* 然后再输出以秒为单位的当前时间值,所以,会打印出4个相同的时间值,如下所示:
* 4:4:1258199615
* 1:1:1258199615
* 3:3:1258199615
* 1:2:1258199615
* 请修改代码,如果有几个线程调用TestDo.doSome(key, value)方法时,传递进去的key相等(equals比较为true),
* 则这几个线程应互斥排队输出结果,即当有两个线程的key都是"1"时,它们中的一个要比另外其他线程晚1秒输出结果,如下所示:
* 4:4:1258199615
* 1:1:1258199615
* 3:3:1258199615
* 1:2:1258199616
* 总之,当每个线程中指定的key相等时,这些相等key的线程应每隔一秒依次输出时间值(要用互斥),
* 如果key不同,则并行执行(相互之间不互斥)
*
*/
//不能改动此Test类
public class Test extends Thread {
private TestDo testDo;
private String key;
private String value;
public Test(String key, String key2, String value) {
this.testDo = TestDo.getInstance();
/*
* 常量"1"和"1"是同一个对象,下面这行代码就是要用"1"+""的方式产生新的对象,
* 以实现内容没有改变,仍然相等(都还为"1"),但对象却不再是同一个的效果
*/
this.key = key + key2;
this.value = value;
}
public static void main(String[] args) throws InterruptedException {
Test a = new Test("1", "", "1");
Test b = new Test("1", "", "2");
Test c = new Test("3", "", "3");
Test d = new Test("4", "", "4");
System.out.println("begin:" + (System.currentTimeMillis() / 1000));
a.start();
b.start();
c.start();
d.start();
}
public void run() {
testDo.doSome(key, value);
}
}
class TestDo {
private TestDo() {
}
private static TestDo _instance = new TestDo();
public static TestDo getInstance() {
return _instance;
}
public void doSome(Object key, String value) {
// 以大括号内的是需要局部同步的代码,不能改动!
{
try {
Thread.sleep(1000);
System.out.println(key + ":" + value + ":"
+ (System.currentTimeMillis() / 1000));
} catch (InterruptedException e) {
e.printStackTrace();
}
}
}
}
运行结果
begin:1374237680
1:2:1374237681
4:4:1374237681
3:3:1374237681
1:1:1374237681
修改:
package syn;
import java.util.ArrayList;
import java.util.Iterator;
import java.util.concurrent.CopyOnWriteArrayList;
//不能改动此Test类
public class Test extends Thread{
private TestDo testDo;
private String key;
private String value;
public Test(String key,String key2,String value){
this.testDo = TestDo.getInstance();
/*常量"1"和"1"是同一个对象,下面这行代码就是要用"1"+""的方式产生新的对象,
以实现内容没有改变,仍然相等(都还为"1"),但对象却不再是同一个的效果*/
this.key = key+key2;
this.value = value;
}
public static void main(String[] args) throws InterruptedException{
Test a = new Test("1","","1");
Test b = new Test("1","","2");
Test c = new Test("3","","3");
Test d = new Test("4","","4");
System.out.println("begin:"+(System.currentTimeMillis()/1000));
a.start();
b.start();
c.start();
d.start();
}
public void run(){
testDo.doSome(key, value);
}
}
class TestDo {
private TestDo() {}
private static TestDo _instance = new TestDo();
public static TestDo getInstance() {
return _instance;
}
//private ArrayList keys = new ArrayList();//线程不安全
private CopyOnWriteArrayList keys = new CopyOnWriteArrayList();
public void doSome(Object key, String value) {
Object o = key;
if(!keys.contains(o)){
keys.add(o);
}else{
for(Iterator iter=keys.iterator();iter.hasNext();){
try {
Thread.sleep(20);
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
Object oo = iter.next();
if(oo.equals(o)){
o = oo;
break;
}
}
}
synchronized(o)
// 以大括号内的是需要局部同步的代码,不能改动!
{
try {
Thread.sleep(1000);
System.out.println(key+":"+value + ":"
+ (System.currentTimeMillis() / 1000));
} catch (InterruptedException e) {
e.printStackTrace();
}
}
}
}
运行结果
begin:1374237873
1:2:1374237874
4:4:1374237874
3:3:1374237874
1:1:1374237875
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