某天，在某个公众号上看到大佬的一份面经，里边出现了这道题.
原题描述是这样的:
使用两个线程打印数字1到100。要求：一个线程打印奇数，另一个打印偶数，并且按照1到100从小到大的顺序正确打印。
下面直接上代码，给出我的实现:
1. 使用Lock实现
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| package me.chan;
import java.util.concurrent.locks.ReentrantLock;
public class Sout1To100 {
private static final ReentrantLock lock = new ReentrantLock();
private static volatile int num = 1;
private static volatile boolean flag = false;
public static void main(String[] args) throws InterruptedException {
Thread t1 = new Thread(()->{
while (num <= 100) {
if (flag) {
try {
lock.lock();
System.out.println(Thread.currentThread().getName()+":"+num);
++num;
flag = false;
} finally {
lock.unlock();
}
}
}
});
Thread t2 = new Thread(()->{
while (num <= 100) {
if (!flag) {
try {
lock.lock();
System.out.println(Thread.currentThread().getName()+":"+num);
++num;
flag = true;
} finally {
lock.unlock();
}
}
}
});
t1.start();
t2.start();
t1.join();
t2.join();
}
|
2. 使用notify()和wait()实现
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| package me.chan;
public class Sout1To100 {
private static final Object lock = new Object();
private static volatile int num = 1;
private static volatile boolean flag = true;
public static void main(String[] args) throws InterruptedException {
Thread t1 = new Thread(()->{
while (num <= 100) {
if (flag) {
synchronized(lock) {
lock.notify();
System.out.println(Thread.currentThread().getName()+":"+num);
++num;
try {
flag = false;
lock.wait(10);
} catch (Exception e) {
e.printStackTrace();
}
}
}
}
});
Thread t2 = new Thread(()->{
while (num <= 100) {
if (!flag) {
synchronized(lock) {
lock.notify();
System.out.println(Thread.currentThread().getName()+":"+num);
++num;
try {
flag = true;
lock.wait(10);
} catch (Exception e) {
e.printStackTrace();
}
}
}
}
});
t1.start();
t2.start();
t1.join();
t2.join();
}
}
|
