LeetCode From Easy To Hard No7[Medium]: Print Zero Even Odd 按顺序打印[010203...0n]

本文链接：https://blog.youkuaiyun.com/code_me_now/article/details/98060208

本文介绍了一道Medium难度的编程题，旨在通过多线程实现特定序列的输出。利用Semaphore来同步三个线程，确保按010203...的顺序输出数字，长度为2n。示例代码展示了如何在Java中优雅地解决这一问题。

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第一次挑战Medium的题：

Suppose you are given the following code:

class ZeroEvenOdd {
  public ZeroEvenOdd(int n) { ... }      // constructor
  public void zero(printNumber) { ... }  // only output 0's
  public void even(printNumber) { ... }  // only output even numbers
  public void odd(printNumber) { ... }   // only output odd numbers
}

The same instance of ZeroEvenOdd will be passed to three different threads:

Thread A will call zero() which should only output 0's.
Thread B will call even() which should only ouput even numbers.
Thread C will call odd() which should only output odd numbers.

Each of the thread is given a printNumber method to output an integer. Modify the given program to output the series 010203040506... where the length of the series must be 2n.

Example 1:

Input: n = 2
Output: "0102"
Explanation: There are three threads being fired asynchronously. One of them calls zero(), the other calls even(), and the last one calls odd(). "0102" is the correct output.

Example 2:

Input: n = 5
Output: "0102030405"

意思就是要我们输出【0102030405...0n】

之前学会了使用Semaphore，刚好这道题可以用得上：

import java.util.concurrent.Semaphore;
import java.util.function.IntConsumer;

class ZeroEvenOdd {

    private int n;

    Semaphore zeroLock, oddLock, evenLock;

    public ZeroEvenOdd(int n) {
        this.n = n;
        this.zeroLock = new Semaphore(1);
        this.oddLock = new Semaphore(0);
        this.evenLock = new Semaphore(0);
    }

    // printNumber.accept(x) outputs "x", where x is an integer.
    public void zero(IntConsumer printNumber) throws InterruptedException {
        int mark = 0;
        while (mark < n) {
            zeroLock.acquire();
            printNumber.accept(0);
            mark ++;
            if (mark % 2 == 0) {
                evenLock.release();
            } else {
                oddLock.release();
            }
        }
    }

    public void even(IntConsumer printNumber) throws InterruptedException {
        int even = 2;
        while (even <= n) {
            evenLock.acquire();
            printNumber.accept(even);
            even += 2;
            zeroLock.release();
        }
    }

    public void odd(IntConsumer printNumber) throws InterruptedException {
        int odd = 1;
        while (odd <= n) {
            oddLock.acquire();
            printNumber.accept(odd);
            odd +=2;
            zeroLock.release();
        }
    }
}