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在多线程并发请求的业务场景中，如何让指定的顺序按需加载类或执行业务逻辑。下面通过对AQS包中的AtomicInteger、CountDownLatch，Semaphore的实际应用来解决按照一定的顺序执行问题。
题目描述
我们提供了一个类：
public class Foo {
public void first() { print("first"); }
public void second() { print("second"); }
public void third() { print("third"); }
}
三个不同的线程 A、B、C 将会共用一个 Foo 实例。
一个将会调用 first() 方法
一个将会调用 second() 方法
还有一个将会调用 third() 方法
请设计修改程序，以确保 second() 方法在 first() 方法之后被执行，third() 方法在 second() 方法之后被执行。
题目示例
题目解法
解法一：AtomicInteger模拟锁的存在
原子性操作和锁的实现
/**
* AtomicInter,累加器
*/
class Foo {
private final AtomicInteger firstAtomic = new AtomicInteger(0);
private final AtomicInteger secondAtomic = new AtomicInteger(0);
public Foo() {
}
public void first(Runnable printFirst) throws InterruptedException {
// printFirst.run() outputs "first". Do not change or remove this line.
printFirst.run();
firstAtomic.incrementAndGet();
}
public void second(Runnable printSecond) throws InterruptedException {
while(firstAtomic.get() != 1){
// wait first
}
// printSecond.run() outputs "second". Do not change or remove this line.
printSecond.run();
secondAtomic.incrementAndGet();
}
public void third(Runnable printThird) throws InterruptedException {
while(secondAtomic.get() != 1){
// wait second
}
// printThird.run() outputs "third". Do not change or remove this line.
printThird.run();
}
}
解法二：使用CountDownLatch倒计时
倒时技术的工具类CountDownLatch
/**
* CountDownLatch
*/
class Foo {
private final CountDownLatch firstCountDown = new CountDownLatch(1);
private final CountDownLatch secondCountDown = new CountDownLatch(1);
public Foo() {
}
public void first(Runnable printFirst) throws InterruptedException {
// printFirst.run() outputs "first". Do not change or remove this line.
printFirst.run();
firstCountDown.countDown();
}
public void second(Runnable printSecond) throws InterruptedException {
firstCountDown.await();
// printSecond.run() outputs "second". Do not change or remove this line.
printSecond.run();
secondCountDown.countDown();
}
public void third(Runnable printThird) throws InterruptedException {
secondCountDown.await();
// printThird.run() outputs "third". Do not change or remove this line.
printThird.run();
}
}
解法三：Semaphore代表锁出战
使用信号量显示执行锁操作
/**
* Semapure信号量
*/
class Foo {
// 初始值为0，用于需要时release
private final Semaphore secondSem = new Semaphore(0);
private final Semaphore thirdSem = new Semaphore(0);
public Foo() {
}
public void first(Runnable printFirst) throws InterruptedException {
// printFirst.run() outputs "first". Do not change or remove this line.
printFirst.run();
secondSem.release();
}
public void second(Runnable printSecond) throws InterruptedException {
// 等待2号信号获取非空信号量
secondSem.acquire();
// printSecond.run() outputs "second". Do not change or remove this line.
printSecond.run();
thirdSem.release();
}
public void third(Runnable printThird) throws InterruptedException {
// 等待3号信号获取非空信号量
thirdSem.acquire();
// printThird.run() outputs "third". Do not change or remove this line.
printThird.run();
}
}
LeetCode原题链接：1114. 按序打印 - 力扣（LeetCode） (leetcode-cn.com)
