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();
}
}