LeetCode 1195. Fizz Buzz Multithreaded 解题报告（python）

1195. Fizz Buzz Multithreaded

1195. Fizz Buzz Multithreaded python solution

题目描述

Write a program that outputs the string representation of numbers from 1 to n, however:

If the number is divisible by 3, output “fizz”.
If the number is divisible by 5, output “buzz”.
If the number is divisible by both 3 and 5, output “fizzbuzz”.
For example, for n = 15, we output: 1, 2, fizz, 4, buzz, fizz, 7, 8, fizz, buzz, 11, fizz, 13, 14, fizzbuzz.

Suppose you are given the following code:

class FizzBuzz {
public FizzBuzz(int n) { … } // constructor
public void fizz(printFizz) { … } // only output “fizz”
public void buzz(printBuzz) { … } // only output “buzz”
public void fizzbuzz(printFizzBuzz) { … } // only output “fizzbuzz”
public void number(printNumber) { … } // only output the numbers
}
Implement a multithreaded version of FizzBuzz with four threads. The same instance of FizzBuzz will be passed to four different threads:

Thread A will call fizz() to check for divisibility of 3 and outputs fizz.
Thread B will call buzz() to check for divisibility of 5 and outputs buzz.
Thread C will call fizzbuzz() to check for divisibility of 3 and 5 and outputs fizzbuzz.
Thread D will call number() which should only output the numbers.

解析

学习一下多线程编程

// An highlighted block
from threading import Lock
class FizzBuzz(object):
    def __init__(self, n):
        self.n = n
        self.c = 1
        
        self.l1 = Lock()
        self.l2 = Lock()
        self.l3 = Lock()
        self.l4 = Lock()
        
        self.l1.acquire()
        self.l2.acquire()
        self.l3.acquire()
        #self.l4.acquire()
        

    # printFizz() outputs "fizz"
    def fizz(self, printFizz):
        """
        :type printFizz: method
        :rtype: void
        """
        
        self.func(printFizz, self.l1)
    	
    # printBuzz() outputs "buzz"
    def buzz(self, printBuzz):
        """
        :type printBuzz: method
        :rtype: void
        """
        
        self.func(printBuzz, self.l2)
    	
    # printFizzBuzz() outputs "fizzbuzz"
    def fizzbuzz(self, printFizzBuzz):
        """
        :type printFizzBuzz: method
        :rtype: void
        """
        self.func(printFizzBuzz, self.l3)
        
    # printNumber(x) outputs "x", where x is an integer.
    def number(self, printNumber):
        """
        :type printNumber: method
        :rtype: void
        """
        while True:
            self.l4.acquire()
            if self.c >= self.n+1:
                break
            printNumber(self.c)
            self.release()
        self.l4.release()
    
    def func(self, f, l):
        while True:
            l.acquire()
            if self.c >= self.n+1:
                break
            f()
            self.release()
        l.release()

    def release(self):
        self.c+=1
        if self.c >= self.n+1:
            self.l1.release()
            self.l2.release()
            self.l3.release()
            self.l4.release()
        elif self.c%15==0:
            self.l3.release()
        elif self.c%3==0:
            self.l1.release()
        elif self.c%5==0:
            self.l2.release()
        else:
            self.l4.release()

Reference

https://leetcode.com/problems/fizz-buzz-multithreaded/discuss/416185/Python-100-100-using-threading.Lock