多线程编程解决思路

场景一，多个线程要求按照指定的顺序执行， 如现在有A、B、C 3个线程，要求按照A->B->C的顺序来依次执行。

解决思路： A B C 线程需要共享一个标志位，依据标志位来决定A B C是该立即执行还是该阻塞或者是等待或者是让出时间片。

方案1， 用AtomicInteger作为标识位， 用该变量对线程数（3）取模，如果非特定值，则让当前线程让出CPU时间片。否则执行业务逻辑并对标识位进行更新。

public class SeqPrint {
    private int n;


    public SeqPrint(int n) {
        this.n = n;
    }

    AtomicLong idx = new AtomicLong(0L);
    public void foo() throws InterruptedException {
        for (int i = 0; i < n; i++) {
            while (idx.get() % 2 != 0) {
                Thread.yield();
            }
            System.out.print("foo");
            idx.getAndIncrement();
        }
    }

    public void bar() throws InterruptedException {
        for (int i = 0; i < n; i++) {
            while (idx.get() % 2 == 0) {
                Thread.yield();
            }
            System.out.print("bar");
            idx.getAndIncrement();
        }
    }
   public static void main(String[] args) throws InterruptedException {
        final SeqPrint seqPrint = new SeqPrint(10);

        Thread t1 = new Thread(new Runnable() {
            public void run() {
                try {
                    seqPrint.foo();
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }
            }
        });

        Thread t2 = new Thread(new Runnable() {
            public void run() {
                try {
                    seqPrint.bar();
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }
            }
        });

        t2.start();
        t1.start();
    }
}

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