2个线程交替打印大小写英文字母

题目

1. 利用wait和notify函数
2. 利用volatile的可见性（volatile能保证可见性，有序性，不能保证原子性，这个一定要牢牢记住）
3. 利用Exchanger类

解法一

public class Solution {

    private static final Object lock = new Object();
    private static volatile boolean flag = true;

    public static void main(String[] args) throws InterruptedException {
        char[] result = new char[52];
        long totalStart = System.currentTimeMillis();
        Thread thread1 = new Thread(() -> {
            long thread1Start = System.currentTimeMillis();
            for (int i = 0; i < 26; i++) {
                synchronized (lock) {
                    if (flag) {
                        result[i * 2] = (char)('a' + i);
                        flag = false;
                        lock.notify();
                    } else {
                        try {
                            lock.wait();
                        } catch (InterruptedException e) {
                            e.printStackTrace();
                        }
                    }
                }
            }
            long thread1Cost = System.currentTimeMillis() - thread1Start;
            System.out.println("thread1Cost " + thread1Cost);
        });
        Thread thread2 = new Thread(() -> {
            long thread2Start = System.currentTimeMillis();
            for (int i = 0; i < 26; i++) {
                synchronized (lock) {
                    if (!flag) {
                        result[i * 2 + 1] = (char)('A' + i);
                        flag = true;
                        lock.notify();
                    } else {
                        if (i != 25) {
                            try {
                                lock.wait();
                            } catch (InterruptedException e) {
                                e.printStackTrace();
                            }
                        }
                    }
                }
            }
            long thread2Cost = System.currentTimeMillis() - thread2Start;
            System.out.println("thread2Cost " + thread2Cost);
        });
        thread1.start();
        thread2.start();
        thread1.join();
        thread2.join();
        // aAbBcCdDeEfFgGhHiIjJkKlLmMnNoOpPqQrRsStTuUvVwWxXyYzZ
        System.out.println(result);
        long totalCost = System.currentTimeMillis() - totalStart;
        // totalCost 119
        System.out.println("totalCost " + totalCost);
    }
}

解法二

public class Solution {

    private static volatile boolean flag = true;

    public static void main(String[] args) throws InterruptedException {
        char[] result = new char[52];
        long totalStart = System.currentTimeMillis();
        Thread thread1 = new Thread(() -> {
            long thread1Start = System.currentTimeMillis();
            for (int i = 0; i < 26;) {
                if (flag) {
                    result[i * 2] = (char)('a' + i);
                    flag = false;
                    i++;
                }
            }
            long thread1Cost = System.currentTimeMillis() - thread1Start;
            System.out.println("thread1Cost " + thread1Cost);
        });
        Thread thread2 = new Thread(() -> {
            long thread2Start = System.currentTimeMillis();
            for (int i = 0; i < 26;) {
                if (!flag) {
                    result[i * 2 + 1] = (char)('A' + i);
                    flag = true;
                    i++;
                }
            }
            long thread2Cost = System.currentTimeMillis() - thread2Start;
            System.out.println("thread2Cost " + thread2Cost);
        });
        thread1.start();
        thread2.start();
        thread1.join();
        thread2.join();
        // aAbBcCdDeEfFgGhHiIjJkKlLmMnNoOpPqQrRsStTuUvVwWxXyYzZ
        System.out.println(result);
        long totalCost = System.currentTimeMillis() - totalStart;
        // totalCost 122
        System.out.println("totalCost " + totalCost);
    }
}

解法三