java n 个线程顺序打印 n 个字符的通用解决方案

闲来无事，练习一下，主要用重入锁的 condition 解决了 notify 的效率问题

import java.util.concurrent.locks.Condition;
import java.util.concurrent.locks.ReentrantLock;

public class Main {

public static ReentrantLock reentrantLock =new ReentrantLock();

//从某个字符开始
 public static Character next = 'A';

public static void main(String[] args) {

    // input
 Character[] characters = {'A', 'B', 'C', 'D'};

    Integer length = characters.length;

    Condition[] conditions = new Condition[length];
    for (int j = 0; j < length; j++) {
        conditions[j] = reentrantLock.newCondition();
    }

    for (int j = 0; j < length; j++) {
        new Thread(new Irunable(conditions[j], characters[j], characters[j + 1 >= length ? 0 : j + 1], conditions[j + 1 >= length ? 0 : j + 1])).start();
    }

}

}

class Irunable implements Runnable {

public Condition currentCondition;

public Character c;

public Character nextC;

public Condition afterCondition;

public Irunable(Condition currentCondition, Character c, Character nextC, Condition afterCondition) {
    this.currentCondition = currentCondition;
    this.c = c;
    this.nextC = nextC;
    this.afterCondition = afterCondition;
}

@Override
 public void run() {

    while (true) {

        try {

            Main.reentrantLock.lock();

            if (c != Main.next) {
                currentCondition.await();
            }

            Thread.sleep(1000);
            System.out.println(c);
            Main.next = nextC;
            afterCondition.signal();

        } catch (InterruptedException e) {
            e.printStackTrace();
        } finally {
            Main.reentrantLock.unlock();
        }
    }

}
}