Question

1116. Print Zero Even Odd

中文文档

Description

You have a function printNumber that can be called with an integer parameter and prints it to the console.

• For example, calling printNumber(7) prints 7 to the console.

You are given an instance of the class ZeroEvenOdd that has three functions: zero, even, and odd. The same instance of ZeroEvenOdd will be passed to three different threads:

• Thread A: calls zero() that should only output 0's.
• Thread B: calls even() that should only output even numbers.
• Thread C: calls odd() that should only output odd numbers.

Modify the given class to output the series "010203040506..." where the length of the series must be 2n.

Implement the ZeroEvenOdd class:

• ZeroEvenOdd(int n) Initializes the object with the number n that represents the numbers that should be printed.
• void zero(printNumber) Calls printNumber to output one zero.
• void even(printNumber) Calls printNumber to output one even number.
• void odd(printNumber) Calls printNumber to output one odd number.

 

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"

 

Constraints:

• 1 <= n <= 1000

Solutions

Solution 1: Multithreading + Semaphore

We use three semaphores $z$, $e$, and $o$ to control the execution order of the three threads, where $z$ is initially set to $1$, and $e$ and $o$ are set to $0$.

• Semaphore $z$ controls the execution of the zero function. When the value of semaphore $z$ is $1$, the zero function can be executed. After execution, the value of semaphore $z$ is set to $0$, and the value of semaphore $e$ or $o$ is set to $1$, depending on whether the even function or the odd function needs to be executed next.
• Semaphore $e$ controls the execution of the even function. When the value of semaphore $e$ is $1$, the even function can be executed. After execution, the value of semaphore $z$ is set to $1$, and the value of semaphore $e$ is set to $0$.
• Semaphore $o$ controls the execution of the odd function. When the value of semaphore $o$ is $1$, the odd function can be executed. After execution, the value of semaphore $z$ is set to $1$, and the value of semaphore $o$ is set to $0$.

The time complexity is $O(n)$, and the space complexity is $O(1)$.

from threading import Semaphore


class ZeroEvenOdd:
  def __init__(self, n):
    self.n = n
    self.z = Semaphore(1)
    self.e = Semaphore(0)
    self.o = Semaphore(0)

  # printNumber(x) outputs "x", where x is an integer.
  def zero(self, printNumber: 'Callable[[int], None]') -> None:
    for i in range(self.n):
      self.z.acquire()
      printNumber(0)
      if i % 2 == 0:
        self.o.release()
      else:
        self.e.release()

  def even(self, printNumber: 'Callable[[int], None]') -> None:
    for i in range(2, self.n + 1, 2):
      self.e.acquire()
      printNumber(i)
      self.z.release()

  def odd(self, printNumber: 'Callable[[int], None]') -> None:
    for i in range(1, self.n + 1, 2):
      self.o.acquire()
      printNumber(i)
      self.z.release()

class ZeroEvenOdd {
  private int n;
  private Semaphore z = new Semaphore(1);
  private Semaphore e = new Semaphore(0);
  private Semaphore o = new Semaphore(0);

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

  // printNumber.accept(x) outputs "x", where x is an integer.
  public void zero(IntConsumer printNumber) throws InterruptedException {
    for (int i = 0; i < n; ++i) {
      z.acquire(1);
      printNumber.accept(0);
      if (i % 2 == 0) {
        o.release(1);
      } else {
        e.release(1);
      }
    }
  }

  public void even(IntConsumer printNumber) throws InterruptedException {
    for (int i = 2; i <= n; i += 2) {
      e.acquire(1);
      printNumber.accept(i);
      z.release(1);
    }
  }

  public void odd(IntConsumer printNumber) throws InterruptedException {
    for (int i = 1; i <= n; i += 2) {
      o.acquire(1);
      printNumber.accept(i);
      z.release(1);
    }
  }
}

#include <semaphore.h>

class ZeroEvenOdd {
private:
  int n;
  sem_t z, e, o;

public:
  ZeroEvenOdd(int n) {
    this->n = n;
    sem_init(&z, 0, 1);
    sem_init(&e, 0, 0);
    sem_init(&o, 0, 0);
  }

  // printNumber(x) outputs "x", where x is an integer.
  void zero(function<void(int)> printNumber) {
    for (int i = 0; i < n; ++i) {
      sem_wait(&z);
      printNumber(0);
      if (i % 2 == 0) {
        sem_post(&o);
      } else {
        sem_post(&e);
      }
    }
  }

  void even(function<void(int)> printNumber) {
    for (int i = 2; i <= n; i += 2) {
      sem_wait(&e);
      printNumber(i);
      sem_post(&z);
    }
  }

  void odd(function<void(int)> printNumber) {
    for (int i = 1; i <= n; i += 2) {
      sem_wait(&o);
      printNumber(i);
      sem_post(&z);
    }
  }
};