按序打印

1114. 按序打印

给你一个类：

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。

线程 A 将会调用 first() 方法
线程 B 将会调用 second() 方法
线程 C 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。

提示：

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 1：

输入：nums = [1,2,3]
输出："firstsecondthird"
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 first() 方法，线程 B 将会调用 second() 方法，线程 C 将会调用 third() 方法。正确的输出是 "firstsecondthird"。

示例 2：

输入：nums = [1,3,2]
输出："firstsecondthird"
解释：
输入 [1,3,2] 表示线程 A 将会调用 first() 方法，线程 B 将会调用 third() 方法，线程 C 将会调用 second() 方法。正确的输出是 "firstsecondthird"。

提示：

nums 是 [1, 2, 3] 的一组排列

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上次编辑到这里，代码来自缓存点击恢复默认模板

class Foo { public: Foo() { } void first(function<void()> printFirst) { // printFirst() outputs "first". Do not change or remove this line. printFirst(); } void second(function<void()> printSecond) { // printSecond() outputs "second". Do not change or remove this line. printSecond(); } void third(function<void()> printThird) { // printThird() outputs "third". Do not change or remove this line. printThird(); } };

java 解法, 执行用时: 10 ms, 内存消耗: 40.4 MB, 提交时间: 2023-10-11 10:37:10

// 使用同步原语
public class Foo {
    private volatile int flag = 1;
    private final Object object = new Object();

    public Foo() {

    }

    public void first(Runnable printFirst) throws InterruptedException {
        synchronized (object) {
            while (flag != 1) object.wait();
            printFirst.run();
            flag = 2;
            object.notifyAll();
        }
    }

    public void second(Runnable printSecond) throws InterruptedException {
        synchronized (object) {
            while (flag != 2) object.wait();
            printSecond.run();
            flag = 3;
            object.notifyAll();
        }
    }

    public void third(Runnable printThird) throws InterruptedException {
        synchronized (object) {
            while (flag != 3) object.wait();
        }
        printThird.run();
    }
}

cpp 解法, 执行用时: 88 ms, 内存消耗: 7.3 MB, 提交时间: 2023-10-11 10:35:18

#include <semaphore.h>

class Foo {

protected:
    sem_t firstJobDone;
    sem_t secondJobDone;

public:

    Foo() {
        sem_init(&firstJobDone, 0, 0);
        sem_init(&secondJobDone, 0, 0);
    }

    void first(function<void()> printFirst) {
        // printFirst() outputs "first".
        printFirst();
        sem_post(&firstJobDone);
    }

    void second(function<void()> printSecond) {
        sem_wait(&firstJobDone);
        // printSecond() outputs "second".
        printSecond();
        sem_post(&secondJobDone);
        
    }

    void third(function<void()> printThird) {
        sem_wait(&secondJobDone);
        // printThird() outputs "third".
        printThird();
    }
};

java 解法, 执行用时: 10 ms, 内存消耗: 40.2 MB, 提交时间: 2023-10-11 10:34:56

class Foo {

  private AtomicInteger firstJobDone = new AtomicInteger(0);
  private AtomicInteger secondJobDone = new AtomicInteger(0);

  public Foo() {}

  public void first(Runnable printFirst) throws InterruptedException {
    // printFirst.run() outputs "first".
    printFirst.run();
    // mark the first job as done, by increasing its count.
    firstJobDone.incrementAndGet();
  }

  public void second(Runnable printSecond) throws InterruptedException {
    while (firstJobDone.get() != 1) {
      // waiting for the first job to be done.
    }
    // printSecond.run() outputs "second".
    printSecond.run();
    // mark the second as done, by increasing its count.
    secondJobDone.incrementAndGet();
  }

  public void third(Runnable printThird) throws InterruptedException {
    while (secondJobDone.get() != 1) {
      // waiting for the second job to be done.
    }
    // printThird.run() outputs "third".
    printThird.run();
  }
}

python3 解法, 执行用时: 68 ms, 内存消耗: 16 MB, 提交时间: 2023-10-11 10:33:53

# 使用线程锁，2把锁
from threading import Lock

class Foo:
    def __init__(self):
        self.firstJobDone = Lock()
        self.secondJobDone = Lock()
        self.firstJobDone.acquire()
        self.secondJobDone.acquire()

    def first(self, printFirst: 'Callable[[], None]') -> None:
        # printFirst() outputs "first".
        printFirst()
        # Notify the thread that is waiting for the first job to be done.
        self.firstJobDone.release()

    def second(self, printSecond: 'Callable[[], None]') -> None:
        # Wait for the first job to be done
        with self.firstJobDone:
            # printSecond() outputs "second".
            printSecond()
            # Notify the thread that is waiting for the second job to be done.
            self.secondJobDone.release()

    def third(self, printThird: 'Callable[[], None]') -> None:
        # Wait for the second job to be done.
        with self.secondJobDone:
            # printThird() outputs "third".
            printThird()

java 解法, 执行用时: 12 ms, 内存消耗: 38.6 MB, 提交时间: 2020-07-31 00:18:17

import java.util.concurrent.CountDownLatch;

class Foo {
    // 计数器通过使用锁（共享锁、排它锁）实现
    private CountDownLatch second = new CountDownLatch(1);
    private CountDownLatch third = 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();
        second.countDown();
    }

    public void second(Runnable printSecond) throws InterruptedException {
        second.await();
        // printSecond.run() outputs "second". Do not change or remove this line.
        printSecond.run();
        third.countDown();
    }

    public void third(Runnable printThird) throws InterruptedException {
        third.await();
        // printThird.run() outputs "third". Do not change or remove this line.
        printThird.run();
    }
}

java 解法, 执行用时: 13 ms, 内存消耗: 38.1 MB, 提交时间: 2020-07-31 00:11:46

import java.util.concurrent.Semaphore;
class Foo {
    // 定义两个信号量, 分别是1->2,  2->3; 通过信号量的 获取-释放 控制流程
    public Semaphore seam_first_two = new Semaphore(0);
    public Semaphore seam_two_second = 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();
        seam_first_two.release();
    }

    public void second(Runnable printSecond) throws InterruptedException {
        seam_first_two.acquire();
        // printSecond.run() outputs "second". Do not change or remove this line.
        printSecond.run();
        seam_two_second.release();
    }

    public void third(Runnable printThird) throws InterruptedException {
        seam_two_second.acquire();
        // printThird.run() outputs "third". Do not change or remove this line.
        printThird.run();
    }
}

python3 解法, 执行用时: 2292 ms, 内存消耗: 13.8 MB, 提交时间: 2020-07-31 00:04:18

class Foo:
    def __init__(self):
        self.step = 1


    def first(self, printFirst: 'Callable[[], None]') -> None:
        # printFirst() outputs "first". Do not change or remove this line.
        printFirst()
        self.step = 2

    def second(self, printSecond: 'Callable[[], None]') -> None:
        while self.step != 2:
            pass
        # printSecond() outputs "second". Do not change or remove this line.
        printSecond()
        self.step = 3


    def third(self, printThird: 'Callable[[], None]') -> None:
        while self.step != 3:
            pass
        # printThird() outputs "third". Do not change or remove this line.
        printThird()

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