[雪峰磁針石部落格]Python經典面試題:用3種方法實現堆疊和佇列並示例實際應用場景

書籍尋找發表於2019-01-28
Python面試題佇列

介紹

資料結構在計算機中組織儲存，以便我們可以有效地訪問和更改資料。 堆疊和佇列是電腦科學中定義的最早的資料結構。

堆疊

遵循後進先出 (Last-in-First-Out LIFO)原則。

  • push – 在堆疊頂部新增元素：

圖片.png

  • pop – 刪除堆疊頂部的元素：

圖片.png

佇列

遵循先入先出(FIFO:First-in-First-Out)原則。

  • enqueue – 在佇列的開頭新增元素：

圖片.png

  • dequeue – 刪除佇列開頭的元素：

圖片.png

使用列表實現堆疊和佇列

Python的內建List資料結構k堆疊和佇列操作的方法。

堆疊

letters = []

# Let`s push some letters into our list
letters.append(`c`)  
letters.append(`a`)  
letters.append(`t`)  
letters.append(`g`)

# Now let`s pop letters, we should get `g`
last_item = letters.pop()  
print(last_item)

# If we pop again we`ll get `t`
last_item = letters.pop()  
print(last_item)

# `c` and `a` remain
print(letters) # [`c`, `a`]

執行結果

g
t
[`c`, `a`]

佇列

fruits = []

# Let`s enqueue some fruits into our list
fruits.append(`banana`)  
fruits.append(`grapes`)  
fruits.append(`mango`)  
fruits.append(`orange`)

# Now let`s dequeue our fruits, we should get `banana`
first_item = fruits.pop(0)  
print(first_item)

# If we dequeue again we`ll get `grapes`
first_item = fruits.pop(0)  
print(first_item)

# `mango` and `orange` remain
print(fruits) # [`c`, `a`]

執行結果

banana
grapes
[`mango`, `orange`]

使用Deque庫的堆疊和佇列

deque是Double Ended Queue的縮寫 – 可以獲取儲存的第一個或最後一個元素的通用佇列，下面我們使用Deque庫的堆疊和佇列：

from collections import deque

# you can initialize a deque with a list 
numbers = deque()

# Use append like before to add elements
numbers.append(99)  
numbers.append(15)  
numbers.append(82)  
numbers.append(50)  
numbers.append(47)

# You can pop like a stack
last_item = numbers.pop()  
print(last_item) # 47  
print(numbers) # deque([99, 15, 82, 50])

# You can dequeue like a queue
first_item = numbers.popleft()  
print(first_item) # 99  
print(numbers) # deque([15, 82, 50])

執行結果

47
deque([99, 15, 82, 50])
99
deque([15, 82, 50])

參考資料

更嚴格的實現

建立撤消功能 – 允許使用者回溯他們的操作，直到會話開始。堆疊是這種情況的理想選擇。 我們可以通過將其推送到堆疊來記錄使用者所採取的每個操作。 當使用者想要撤消操作時，他們將從堆疊中彈出它。

遊戲中，每次按下按鈕，都會觸發輸入事件。 測試人員注意到，如果按鈕按下得太快，遊戲只處理第一個按鈕，特殊動作將無效！可以使用佇列修復它。 我們可以將所有輸入事件排入佇列。

#!/usr/bin/python3
# -*- coding: utf-8 -*-
# 專案實戰討論QQ群630011153 144081101
# python測試開發庫彙總: https://github.com/china-testing/python-api-tesing/
# 本文最佳板式地址： https://www.jianshu.com/p/c990427ca608
# A simple class stack that only allows pop and push operations
class Stack:

    def __init__(self):
        self.stack = []

    def pop(self):
        if len(self.stack) < 1:
            return None
        return self.stack.pop()

    def push(self, item):
        self.stack.append(item)

    def size(self):
        return len(self.stack)

# And a queue that only has enqueue and dequeue operations
class Queue:

    def __init__(self):
        self.queue = []

    def enqueue(self, item):
        self.queue.append(item)

    def dequeue(self):
        if len(self.queue) < 1:
            return None
        return self.queue.pop(0)

    def size(self):
        return len(self.queue) 
    
document_actions = Stack()

# The first enters the title of the document
document_actions.push(`action: enter; text_id: 1; text: This is my favourite document`)  
# Next they center the text
document_actions.push(`action: format; text_id: 1; alignment: center`)  
# As with most writers, the user is unhappy with the first draft and undoes the center alignment
document_actions.pop()  
# The title is better on the left with bold font
document_actions.push(`action: format; text_id: 1; style: bold`) 

input_queue = Queue()

# The player wants to get the upper hand so pressing the right combination of buttons quickly
input_queue.enqueue(`DOWN`)  
input_queue.enqueue(`RIGHT`)  
input_queue.enqueue(`B`)

# Now we can process each item in the queue by dequeueing them
key_pressed = input_queue.dequeue() # `DOWN`

# We`ll probably change our player position
key_pressed = input_queue.dequeue() # `RIGHT`

# We`ll change the player`s position again and keep track of a potential special move to perform
key_pressed = input_queue.dequeue() # `B`

# This can do the act, but the game`s logic will know to do the special move


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