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| 1 | +package pp.arithmetic.leetcode; |
| 2 | + |
| 3 | +import java.util.concurrent.Semaphore; |
| 4 | +import java.util.function.IntConsumer; |
| 5 | + |
| 6 | +/** |
| 7 | + * Created by wangpeng on 2020-07-09. |
| 8 | + * 1116. 打印零与奇偶数 |
| 9 | + * <p> |
| 10 | + * 假设有这么一个类: |
| 11 | + * <p> |
| 12 | + * class ZeroEvenOdd { |
| 13 | + * public ZeroEvenOdd(int n) { ... } // 构造函数 |
| 14 | + * public void zero(printNumber) { ... } // 仅打印出 0 |
| 15 | + * public void even(printNumber) { ... } // 仅打印出 偶数 |
| 16 | + * public void odd(printNumber) { ... } // 仅打印出 奇数 |
| 17 | + * } |
| 18 | + * 相同的一个 ZeroEvenOdd 类实例将会传递给三个不同的线程: |
| 19 | + * <p> |
| 20 | + * 线程 A 将调用 zero(),它只输出 0 。 |
| 21 | + * 线程 B 将调用 even(),它只输出偶数。 |
| 22 | + * 线程 C 将调用 odd(),它只输出奇数。 |
| 23 | + * 每个线程都有一个 printNumber 方法来输出一个整数。请修改给出的代码以输出整数序列 010203040506... ,其中序列的长度必须为 2n。 |
| 24 | + * <p> |
| 25 | + * |
| 26 | + * <p> |
| 27 | + * 示例 1: |
| 28 | + * <p> |
| 29 | + * 输入:n = 2 |
| 30 | + * 输出:"0102" |
| 31 | + * 说明:三条线程异步执行,其中一个调用 zero(),另一个线程调用 even(),最后一个线程调用odd()。正确的输出为 "0102"。 |
| 32 | + * 示例 2: |
| 33 | + * <p> |
| 34 | + * 输入:n = 5 |
| 35 | + * 输出:"0102030405" |
| 36 | + * <p> |
| 37 | + * 来源:力扣(LeetCode) |
| 38 | + * 链接:https://leetcode-cn.com/problems/print-zero-even-odd |
| 39 | + * 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。 |
| 40 | + */ |
| 41 | +public class _1116_ZeroEvenOdd { |
| 42 | + |
| 43 | + public static void main(String[] args) { |
| 44 | + ZeroEvenOdd zeroEvenOdd = new ZeroEvenOdd(5); |
| 45 | + IntConsumer printNumber = new IntConsumer() { |
| 46 | + @Override |
| 47 | + public void accept(int value) { |
| 48 | + System.out.println(value); |
| 49 | + } |
| 50 | + }; |
| 51 | + new Thread(){ |
| 52 | + @Override |
| 53 | + public void run() { |
| 54 | + super.run(); |
| 55 | + try { |
| 56 | + zeroEvenOdd.zero(printNumber); |
| 57 | + } catch (InterruptedException e) { |
| 58 | + e.printStackTrace(); |
| 59 | + } |
| 60 | + } |
| 61 | + }.start(); |
| 62 | + new Thread(){ |
| 63 | + @Override |
| 64 | + public void run() { |
| 65 | + super.run(); |
| 66 | + try { |
| 67 | + zeroEvenOdd.even(printNumber); |
| 68 | + } catch (InterruptedException e) { |
| 69 | + e.printStackTrace(); |
| 70 | + } |
| 71 | + } |
| 72 | + }.start(); |
| 73 | + new Thread(){ |
| 74 | + @Override |
| 75 | + public void run() { |
| 76 | + super.run(); |
| 77 | + try { |
| 78 | + zeroEvenOdd.odd(printNumber); |
| 79 | + } catch (InterruptedException e) { |
| 80 | + e.printStackTrace(); |
| 81 | + } |
| 82 | + } |
| 83 | + }.start(); |
| 84 | + } |
| 85 | + |
| 86 | + static class ZeroEvenOdd { |
| 87 | + private int n; |
| 88 | + private Semaphore zeroSe = new Semaphore(1); |
| 89 | + private Semaphore evenSe = new Semaphore(0); |
| 90 | + private Semaphore oddSe = new Semaphore(0); |
| 91 | + |
| 92 | + public ZeroEvenOdd(int n) { |
| 93 | + this.n = n; |
| 94 | + } |
| 95 | + |
| 96 | + // printNumber.accept(x) outputs "x", where x is an integer. |
| 97 | + public void zero(IntConsumer printNumber) throws InterruptedException { |
| 98 | + for (int i = 0; i < n; i++) { |
| 99 | + zeroSe.acquire(); |
| 100 | + printNumber.accept(0); |
| 101 | + if (i % 2 == 0) { |
| 102 | + evenSe.release(); |
| 103 | + } else { |
| 104 | + oddSe.release(); |
| 105 | + } |
| 106 | + } |
| 107 | + } |
| 108 | + |
| 109 | + public void even(IntConsumer printNumber) throws InterruptedException { |
| 110 | + for (int i = 1; i <= n; i+=2) { |
| 111 | + evenSe.acquire(); |
| 112 | + printNumber.accept(i); |
| 113 | + zeroSe.release(); |
| 114 | + } |
| 115 | + } |
| 116 | + |
| 117 | + public void odd(IntConsumer printNumber) throws InterruptedException { |
| 118 | + for (int i = 2; i <= n; i+=2) { |
| 119 | + oddSe.acquire(); |
| 120 | + printNumber.accept(i); |
| 121 | + zeroSe.release(); |
| 122 | + } |
| 123 | + } |
| 124 | + } |
| 125 | +} |
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