码迷,mamicode.com
首页 > 编程语言 > 详细

IT十八掌作业_java基础第九天_多线程、自动拆装箱

时间:2016-03-13 18:16:30      阅读:241      评论:0      收藏:0      [点我收藏+]

标签:java基础第九天_多线程、自动拆装箱

1.蜜蜂和熊的生产消费关系,熊在蜂蜜满10斤吃掉。蜜蜂一次生产一斤蜂蜜,且蜜蜂生成一斤蜂蜜花费的时间是10s。

  十只蜜蜂和两只熊。

   class Bee extends Thread{

private int bag;

private static final int BAG_MAX = 20;

private static final int ONCE = 5;

private static final int TIME = 10 ;

private Box box;

private String name;

public Bee(Box box, String name) {

super();

this.box = box;

this.name = name;

}

public void run()

{

//满足放蜂蜜的条件

while (true) {

//如果不足 5

if(bag >= 0 && bag < 5){

bag ++;

try {

System.out.println(name + "加蜂蜜,数量" + bag);

Thread.sleep(10);

} catch (Exception e) {

// TODO: handle exception

}

if (bag >= 5) {

synchronized (box) {

int cap = box.capacity;

if (cap == Box.MAX) {

System.out.println("蜂蜜满了");

try {

box.wait(); // 

} catch (Exception e) {

// TODO: handle exception

}

}//未满

else {

//剩余空间

int remain = Box.MAX - cap ;

if (bag >= remain) {

box.capacity = Box.MAX;

bag = bag - remain;

box.notifyAll();

}

else {

box.capacity = box.capacity + bag;

bag = 0;

System.out.println(name + "加蜂蜜,容器中的数量为" + box.capacity);

}

}

}

}

}

}

}



 class Box {

public static final int MAX = 20;

public int capacity = 0 ;

}




 class Bear extends Thread {

private Box box;

private String name;

public Bear(Box box, String name) {

super();

this.box = box;

this.name = name;

}

public void run()

{

while (true)

{

synchronized (box) {

if (box.capacity > 10)

{

System.out.println("总量为:"+box.capacity);

box.capacity = box.capacity - 10;

System.out.println(name + "吃了,剩余 "+box.capacity);

box.notifyAll();

}else {

try {

System.out.println(name + "等着急了");

box.wait();

} catch (InterruptedException e) {

// TODO Auto-generated catch block

e.printStackTrace();

}

}

}

}

}

}



 class App2 {


public static void main(String[] args) {

// TODO Auto-generated method stub

Box box = new Box();

Bee b1 = new Bee(box,"b1");

Bee b2 = new Bee(box,"b2");

Bee b3 = new Bee(box,"b3");

Bee b4 = new Bee(box,"b4");

Bee b5 = new Bee(box,"b5");

Bee b6 = new Bee(box,"b6");

Bee b7 = new Bee(box,"b7");

Bee b8 = new Bee(box,"b8");

Bee b9 = new Bee(box,"b9");

Bee b10 = new Bee(box,"b10");

Bear bear = new Bear(box,"熊大");

Bear bear2 = new Bear(box,"熊二");

b1.start();

b2.start();

b3.start();

b4.start();

b5.start();

b6.start();

b7.start();

b8.start();

b9.start();

b10.start();

bear.start();

bear2.start();

}


}


2.取出两个字符串中最大的公共子串。

public class StringDemo {


public static void main(String[] args) {

// TODO Auto-generated method stub

String str1 = "abcasdfasdfd";

String str2 = "a13asdfbc3dsadfasd23defsdfs";

int min = 0;

int max = 0;


int len1 = str1.length();

int len2 = str2.length();

//比较连个字符串的长度。

String minStr = len1 <= len2 ? str1 : str2;

String maxStr = len1 > len2 ? str1 : str2;

String targetStr = null;

//循环 从字符串长度小的最大的子串开始

for (int i = minStr.length(); i >= 1; i--) { // 取出子字符串

for (int j = 0; j < minStr.length() - i+1; j++) {

targetStr = minStr.substring(j, j + i);


// 比较子串 和 最大字符串是否匹配

for (int k = 0; k < maxStr.length() - targetStr.length()+1; k++) {

String tmp = maxStr.substring(k, k + targetStr.length());

System.out.println("循环第 " + k + "次");

if (maxStr.substring(k, k + targetStr.length()).equals(targetStr)) {

System.out.println(" 最大子字符串:" + targetStr);

System.exit(-1);

}

}

}

}


}


}



3.StringBuffer是线程安全的,StringBuilder不是线程安全。单线程访问情况下,性能是否一致?

在单线程访问的情况下,性能是一样的。

只有在多线程并发的情况下,StringBuilder的性能高于StringBuffer,

因为StringBuffer有加Synchronized关键字,每次只允许一个线程访问,而StringBuilder是可以同时访问的。


IT十八掌作业_java基础第九天_多线程、自动拆装箱

标签:java基础第九天_多线程、自动拆装箱

原文地址:http://45279.blog.51cto.com/35279/1750519

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!