1116. Print Zero Even Odd

題目敘述

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

題目翻譯

這題的條件是，會有三個 method，分別是 zero, even, odd。會給一個參數 n，然後需要依序從 1 印到 n，但是在每個數字前都還要先印出 0。如果是印出 0 則要執行 zero，印出偶數需要使用 even，奇數則要使用 odd。

解法解析

類似上一題 1115 一樣，使用 for-loop 去處理不同 method 的執行時機。每次 even 或 odd 執行完後，都回到 zero 的狀態。然後在 zero 的 method 中去判斷下次要執行的是 odd 或 even。

程式範例

Python

from threading import Lock


class ZeroEvenOdd:
    def __init__(self, n):
        self.n = n

        self.zero_mutex = Lock()
        self.odd_mutex = Lock()
        self.even_mutex = Lock()

        self.even_mutex.acquire()
        self.odd_mutex.acquire()

    # printNumber(x) outputs "x", where x is an integer.
    def zero(self, printNumber: 'Callable[[int], None]') -> None:
        for i in range(self.n):
            self.zero_mutex.acquire()

            printNumber(0)

            if (i % 2) == 0:
                self.even_mutex.release()
            else:
                self.odd_mutex.release()

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

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

LeetCode Solution, Medium, 1116. Print Zero Even Odd

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1116. Print Zero Even Odd

題目敘述

題目翻譯

解法解析

程式範例

Python

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Comments

LeetCode Solution

More from this blog

1116. Print Zero Even Odd

題目敘述

題目翻譯

解法解析

程式範例

Python