此文首发于我的个人博客：LeetCode 1114. Print in Order–Java解法–并发问题 — zhang0peter的个人博客
LeetCode题解专栏：LeetCode题解
LeetCode 所有题目总结：LeetCode 所有题目总结
大部分题目C++，Python，Java的解法都有。
题目地址：Print in Order - LeetCode
Suppose we have a class:
public class Foo {
public void first() { print("first"); }
public void second() { print("second"); }
public void third() { print("third"); }
}
The same instance of Foo will be passed to three different threads. Thread A will call first(), thread B will call second(), and thread C will call third(). Design a mechanism and modify the program to ensure that second() is executed after first(), and third() is executed after second().
Example 1:
Input: [1,2,3]
Output: "firstsecondthird"
Explanation: There are three threads being fired asynchronously. The input [1,2,3] means thread A calls first(), thread B calls second(), and thread C calls third(). "firstsecondthird" is the correct output.
Example 2:
Input: [1,3,2]
Output: "firstsecondthird"
Explanation: The input [1,3,2] means thread A calls first(), thread B calls third(), and thread C calls second(). "firstsecondthird" is the correct output.
Note:
We do not know how the threads will be scheduled in the operating system, even though the numbers in the input seems to imply the ordering. The input format you see is mainly to ensure our tests’ comprehensiveness.
这道题目的意思是并发时如何确保进程的执行顺序，可以采用信号量。我使用 synchronized, wait, notifyAll实现。
Java解法如下：
class Foo {
private final AtomicInteger i = new AtomicInteger();
private final Object lock = new Object();
public Foo() {
i.set(0);
}
public void first(Runnable printFirst) throws InterruptedException {
synchronized (lock) {
while (i.get() != 0) {
lock.wait();
}
printFirst.run();
i.set(1);
lock.notifyAll();
}
}
public void second(Runnable printSecond) throws InterruptedException {
synchronized (lock) {
while (i.get() != 1) {
lock.wait();
}
printSecond.run();
i.set(2);
lock.notifyAll();
}
}
public void third(Runnable printThird) throws InterruptedException {
synchronized (lock) {
while (i.get() != 2) {
lock.wait();
}
printThird.run();
i.set(3);
}
}
}