交替打印-Interleaving
目录[*]交替打印奇偶数
[*]交替打印ABC
[*]交替打印数字
[*]总结与思考
交替打印奇偶数
题目概述
创建两个线程，一个打印奇数，一个打印偶数。要求两个线程交替输出，顺序打印出 1 2 3 4 5 ... 100。
解法一：synchronized + wait/notify
public class Main {
private static final Object lock = new Object();
private static int count = 1;
public static void main(String[] args) {
Thread odd = new Thread(() -> {
while (count <= 100) {
synchronized (lock) {
if ((count & 1) == 0) {
try {
lock.wait();
} catch (InterruptedException e) {
e.printStackTrace();
}
} else {
System.out.println(Thread.currentThread().getName() + ": " + count);
count++;
lock.notify();
}
}
}
}, "ODD");
Thread even = new Thread(() -> {
while (count <= 100) {
synchronized (lock) {
if ((count & 1) == 1) {
try {
lock.wait();
} catch (InterruptedException e) {
e.printStackTrace();
}
} else {
System.out.println(Thread.currentThread().getName() + ": " + count);
count++;
lock.notify();
}
}
}
}, "EVEN");
odd.start();
even.start();
}
}交替打印ABC
题目概述
创建三个线程，分别打印 "A"、"B"、"C"，要求按顺序交替输出 ABCABC...，共打印 10 次。
解法一：synchronized + wait/notify
import java.util.concurrent.Semaphore;
public class Main {
static int count = 1;
static Semaphore oddSemaphore = new Semaphore(1);
static Semaphore evenSemaphore = new Semaphore(0);
public static void main(String[] args) {
new Thread(() -> {
while (count <= 100) {
try {
oddSemaphore.acquire();
if (count > 100) break;
System.out.println(("ODD : " + count++));
evenSemaphore.release();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
}).start();
new Thread(() -> {
while (count <= 100) {
try {
evenSemaphore.acquire();
if (count > 100) break;
System.out.println(("EVEN : " + count++));
oddSemaphore.release();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
}).start();
}
}解法二：ReentrantLock + Condition
public class Main {
private static int state = 0;
private static final Object lock = new Object();
public static void main(String[] args) {
int loop = 10;
new Thread(() -> printLetter('A', 0, 10)).start();
new Thread(() -> printLetter('B', 1, 10)).start();
new Thread(() -> printLetter('C', 2, 10)).start();
}
private static void printLetter(char ch, int target, int loop) {
for (int i = 0; i < loop;) {
synchronized (lock) {
if (state != target) {
try {
lock.wait();
} catch (InterruptedException e) {
e.printStackTrace();
}
} else {
System.out.println(ch);
state = (state + 1) % 3;
lock.notifyAll();
i++;
}
}
}
}
}交替打印数字
题目概述
用3个线程打印 1~100 的整数，要求3个线程按顺序轮流打印
解法一：synchronized + wait/notify
import java.util.concurrent.locks.*;
public class Main {
private static int state = 0;
private static final Lock lock = new ReentrantLock();
private static final Condition A = lock.newCondition();
private static final Condition B = lock.newCondition();
private static final Condition C = lock.newCondition();
public static void main(String[] args) {
int loop = 10;
new Thread(() -> print('A', 0, A, B, loop)).start();
new Thread(() -> print('B', 1, B, C, loop)).start();
new Thread(() -> print('C', 2, C, A, loop)).start();
}
private static void print(char ch, int target, Condition curr, Condition next, int loop) {
for (int i = 0; i < loop;) {
lock.lock();
try {
while (state != target) {
curr.await();
}
System.out.println(ch);
state = (state + 1) % 3;
i++;
next.signal();
} catch (InterruptedException e) {
e.printStackTrace();
} finally {
lock.unlock();
}
}
}
}解法二：ReentrantLock + Condition
public class Main {
private static int count = 1;
private static final Object lock = new Object();
public static void main(String[] args) {
for (int i = 0; i < 3; i++) {
int id = i;
new Thread(() -> {
while (true) {
synchronized (lock) {
if (count > 100) break;
if (count % 3 != id) {
try {
lock.wait();
} catch (InterruptedException e) {
e.printStackTrace();
}
} else {
System.out.println(count++);
lock.notifyAll();
}
}
}
}).start();
}
}
}总结与思考
[*]用while (true)
[*]及时释放锁避免死锁
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