睡完后什么线程？ 在思考 leetcode 问题时：按顺序打印

Question

I am doing leetcode question: https://leetcode.com/problems/print-in-order/ and try different solution here:

// Solution 1:
// 25 ms time
// The first solution is slower? Probably because waking up a thread is more costly. The second solution is more close to positive lock. 
class Foo {
    
     int lockSecond = 0;
     int lockThird = 0;

    public Foo() {
        
    }

    public synchronized void first(Runnable printFirst) throws InterruptedException {
        
        // printFirst.run() outputs "first". Do not change or remove this line.
        printFirst.run();
        this.lockSecond = 1;
        notifyAll();
    }

    public synchronized void second(Runnable printSecond) throws InterruptedException {
        
        for (;this.lockSecond == 0;) {
            wait();
        }
        
        // printSecond.run() outputs "second". Do not change or remove this line.
        printSecond.run();
        this.lockThird = 1;
        notifyAll();
    }

    public synchronized void third(Runnable printThird) throws InterruptedException {
        for (;this.lockThird == 0;) {
             wait();
        }
        // printThird.run() outputs "third". Do not change or remove this line.
        printThird.run();
        notifyAll();
    }
}
//Solution 2:
//10 ms there is no valotile, which means thread will retieve latest value when finishing sleeping
class Foo {
    
    int lockSecond = 0;
    int lockThird = 0;

    public Foo() {
        
    }

    public void first(Runnable printFirst) throws InterruptedException {
        
        // printFirst.run() outputs "first". Do not change or remove this line.
        printFirst.run();
        this.lockSecond = 1;
    }

    public void second(Runnable printSecond) throws InterruptedException {
        
        for (;this.lockSecond == 0;) {
            Thread.sleep(1);
        }
        
        // printSecond.run() outputs "second". Do not change or remove this line.
        printSecond.run();
        this.lockThird = 1;
    }

    public void third(Runnable printThird) throws InterruptedException {
        for (;this.lockThird==0;) {
           Thread.sleep(1);
        }
        // printThird.run() outputs "third". Do not change or remove this line.
        printThird.run();
    }
}
// same as second solution, 
//Solution 3:
class Foo {
    
    volatile int lockSecond = 0;
    volatile int lockThird = 0;

    public Foo() {
        
    }

    public void first(Runnable printFirst) throws InterruptedException {
        
        // printFirst.run() outputs "first". Do not change or remove this line.
        printFirst.run();
        this.lockSecond = 1;
    }

    public void second(Runnable printSecond) throws InterruptedException {
        
        for (;this.lockSecond == 0;) {
        }
        
        // printSecond.run() outputs "second". Do not change or remove this line.
        printSecond.run();
        this.lockThird = 1;
    }

    public void third(Runnable printThird) throws InterruptedException {
        for (;this.lockThird==0;) {
        }
        // printThird.run() outputs "third". Do not change or remove this line.
        printThird.run();
    }
}

I can understand we use volatile to make the var shared across different thread in solution 3 to avoid the endless loop. If I remove volatile, Leetcode will say exceeding time limit.

But for solution 2, I can remove volatile if I use sleep. Does this mean thread will always re-read value of variable after finishing sleep, which avoid the time limit issue? Not familiar with multi-threading, there might be misleading comments in the code snippet. Welcome any correcting! ----------------------Update --------------------------------

It has nothing to do with sleep() actually. It is up to whether there is statement in the for loop body. If I replace sleep with println(), it still passes test cases. If it is empty, it will exceed time. Actually in my opinion, if there is no valotile keyword. It will get into endless loop since different thread share no visibility but it doesn't. It just gets slower. When there is a statement, it can even pass the tests case of leetcode

Answer 1

Thread.sleep(n) is not guaranteed to do anything except not return until either n milliseconds have elapsed, or the calling thread was interrupted. Anything else that happens when you call it either is a peculiarity of whatever JRE, JVM, and OS version you are using, or else it's a roll of the dice on whatever JRE, JVM, and OS version you are using.

If your code depends on sleep() for any kind of synchronization or coordination between different threads, then you are taking a risk that it won't work everywhere and/or it won't work every time.