1114. Print in Order

題目敘述

Suppose we have a class:

public class Foo {
  public void first() { print("first"); }
  public void second() { print("second"); }
  public void third() { print("third"); }
}

The same instance of Foo will be passed to three different threads. Thread A will call first(), thread B will call second(), and thread C will call third(). Design a mechanism and modify the program to ensure that second() is executed after first(), and third() is executed after second().

Note:

We do not know how the threads will be scheduled in the operating system, even though the numbers in the input seem to imply the ordering. The input format you see is mainly to ensure our tests' comprehensiveness.

Example 1:

Input: [1,2,3]
Output: "firstsecondthird"
Explanation: There are three threads being fired asynchronously. The input [1,2,3] means thread A calls first(), thread B calls second(), and thread C calls third(). "firstsecondthird" is the correct output.

Example 2:

Input: [1,3,2]
Output: "firstsecondthird"
Explanation: The input [1,3,2] means thread A calls first(), thread B calls third(), and thread C calls second(). "firstsecondthird" is the correct output.

Constraints:

nums is a permutation of [1, 2, 3].

題目翻譯

這題很簡單就是需要按照順序去執行 method，按照順序印出字串。

解法解析

這題目就是很依賴程式語言的特性，所以可以發現能使用的程式語言相對的少。這邊使用 Python 的標準套件 threading，這邊使用的是 Lock 來做解題。

程式範例

Python

from threading import Lock

class Foo:
    def __init__(self):
        self.firstJobDone = Lock()
        self.secondJobDone = Lock()
        self.firstJobDone.acquire()
        self.secondJobDone.acquire()

    def first(self, printFirst: 'Callable[[], None]') -> None:
        # printFirst() outputs "first".
        printFirst()
        # Notify the thread that is waiting for the first job to be done.
        self.firstJobDone.release()

    def second(self, printSecond: 'Callable[[], None]') -> None:
        # Wait for the first job to be done
        with self.firstJobDone:
            # printSecond() outputs "second".
            printSecond()
            # Notify the thread that is waiting for the second job to be done.
            self.secondJobDone.release()

    def third(self, printThird: 'Callable[[], None]') -> None:

        # Wait for the second job to be done.
        with self.secondJobDone:
            # printThird() outputs "third".
            printThird()

LeetCode Solution, Easy, 1114. Print in Order

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1114. Print in Order

題目敘述

題目翻譯

解法解析

程式範例

Python

Command Palette

Comments

LeetCode Solution

More from this blog

1114. Print in Order

題目敘述

題目翻譯

解法解析

程式範例

Python