• LeetCode_1114.按顺序打印(多线程)


    LeetCode_1114

    我们提供了一个类:

    public class Foo {
        public void one() { print("one"); }
        public void two() { print("two"); }
        public void three() { print("three"); }
    }
    
    三个不同的线程将会共用一个 Foo 实例。
    
    线程 A 将会调用 one() 方法
    线程 B 将会调用 two() 方法
    线程 C 将会调用 three() 方法
    请设计修改程序,以确保 two() 方法在 one() 方法之后被执行,
    three() 方法在 two() 方法之后被执行。
    

    示例 1:

    输入: [1,2,3]
    输出: "onetwothree"
    解释: 
    有三个线程会被异步启动。
    输入 [1,2,3] 表示线程 A 将会调用 one() 方法,
    线程 B 将会调用 two() 方法,线程 C 将会调用 three() 方法。
    正确的输出是 "onetwothree"。
    

    示例 2:

    输入: [1,3,2]
    输出: "onetwothree"
    解释: 
    输入 [1,3,2] 表示线程 A 将会调用 one() 方法,
    线程 B 将会调用 three() 方法,线程 C 将会调用 two() 方法。
    正确的输出是 "onetwothree"。
    

    注意:

    尽管输入中的数字似乎暗示了顺序,
    但是我们并不保证线程在操作系统中的调度顺序。
    你看到的输入格式主要是为了确保测试的全面性。
    

    示例代码:

    class Foo {
    
        public Foo() {
            
        }
    
        public void first(Runnable printFirst) throws InterruptedException {
            
            // printFirst.run() outputs "first". Do not change or remove this line.
            printFirst.run();
        }
    
        public void second(Runnable printSecond) throws InterruptedException {
            
            // printSecond.run() outputs "second". Do not change or remove this line.
            printSecond.run();
        }
    
        public void third(Runnable printThird) throws InterruptedException {
            
            // printThird.run() outputs "third". Do not change or remove this line.
            printThird.run();
        }
    }
    

    方法一:使用锁题解

    • 测试用例:36个
    • 执行用时:17ms
    • 内存消耗:35.8MB
    class Foo {
    
        // 构造两道屏障
        private boolean firstFinished;
        private boolean secondFinished;
        private Object lock = new Object();
        
        public Foo() {
            
        }
    
        public void first(Runnable printFirst) throws InterruptedException {
            
            synchronized (lock) {
                // printFirst.run() outputs "first". Do not change or remove this line.
                printFirst.run();
                firstFinished = true;
                lock.notifyAll();
            }
        }
    
        public void second(Runnable printSecond) throws InterruptedException {
            
            synchronized (lock) {
                while (!firstFinished) {
                    lock.wait();
                }
                // printSecond.run() outputs "second". Do not change or remove this line.
                printSecond.run();
                secondFinished = true;
                lock.notifyAll();
            }
        }
    
        public void third(Runnable printThird) throws InterruptedException {
            
            synchronized (lock) {
                while (!secondFinished) {
                    lock.wait();
                }
                // printThird.run() outputs "third". Do not change or remove this line.
                printThird.run();
            }
        }
    }
    

    方法二:通过信号量题解

    • 测试用例:36个
    • 执行用时:19ms
    • 内存消耗:36.1MB
    import java.util.concurrent.Semaphore;
    
    class Foo {
        
        Semaphore A;
        Semaphore B;
        Semaphore C;
    
        public Foo() {
            A = new Semaphore(1);
            B = new Semaphore(0);
            C = new Semaphore(0);
        }
    
        public void first(Runnable printFirst) throws InterruptedException {
            
            A.acquire();
            // printFirst.run() outputs "first". Do not change or remove this line.
            printFirst.run();
            B.release();
        }
    
        public void second(Runnable printSecond) throws InterruptedException {
            
            B.acquire();
            // printSecond.run() outputs "second". Do not change or remove this line.
            printSecond.run();
            C.release();
        }
    
        public void third(Runnable printThird) throws InterruptedException {
            
            C.acquire();
            // printThird.run() outputs "third". Do not change or remove this line.
            printThird.run();
        }
    }
    

    方法三:使用 CountDownLatch 题解

    • 测试用例:36个
    • 执行用时:19ms
    • 内存消耗:36MB
    import java.util.concurrent.CountDownLatch;
    
    class Foo {
    
        CountDownLatch latch1;
        CountDownLatch latch2;
        
        public Foo() {
            latch1 = new CountDownLatch(1);
            latch2 = new CountDownLatch(2);
        }
    
        public void first(Runnable printFirst) throws InterruptedException {
            
            // printFirst.run() outputs "first". Do not change or remove this line.
            printFirst.run();
            latch1.countDown();
            latch2.countDown();
        }
    
        public void second(Runnable printSecond) throws InterruptedException {
            
            latch1.await();
            // printSecond.run() outputs "second". Do not change or remove this line.
            printSecond.run();
            latch2.countDown();
        }
    
        public void third(Runnable printThird) throws InterruptedException {
            
            latch2.await();
            // printThird.run() outputs "third". Do not change or remove this line.
            printThird.run();
        }
    }
    
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  • 原文地址:https://www.cnblogs.com/wumz/p/11306502.html
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