Priority Queue Linked List Implementation

Question

I am new to Python and wanted to make sure my code answers the question for my assignment due.

My Task:

A priority queue is a queue in which each item is assigned a priority and items with a higher priority are removed before those with a lower priority, irrespective of when they were added. Integer values are used for the priorities with a smaller integer value having a higher priority. In an unbounded priority queue, there are no limits on the range of integer values that can be used for the priorities.

• PriorityQueue(): Creates a new empty unbounded priority queue.

• isEmpty(): Returns a boolean value indicating whether the queue is empty.

• length(): Returns the number of items currently in the queue.

• enqueue(item, priority): Adds the given item to the queue by inserting it in the proper position based on the given priority.

• dequeue(): Removes and returns the next item from the queue, which is the item with the highest priority. If two or more items have the same priority, those items are removed in FIFO order. An item cannot be dequeued from an empty queue.

• peek(): Returns a copy of the next item in the queue, without removing the item. The next item is the same value that would be returned by the dequeue operation. An item cannot be dequeued from an empty queue.

My Code:

class Node( object ) :
 def __init__( self, cargo = None, next = None ) :
 self.cargo = cargo
 self.next = next
 # Creates a new empty unbounded priority queue
class PriorityQueue :
 # 
 def __init__( self ) :
 self.length = 0
 self.head = None
 self.last = None
 # Returns a boolean value indicating whether the queue is empty
 def isEmpty( self ) :
 return (self.length == 0)
 # Returns the number of items currently in the queue
 def __len__( self ) :
 return len(self.length)
 # Adds the given item to the queue by inserting it in the proper position 
 # based on the given priority. The new node is appeneded to the end of the
 # linked list
 def enqueue( self, item, priority) :
 newNode = Node(cargo)
 newNode.next = None
 if self.length == 0:
 self.head self.last = newNode
 newNode.next = self.head
 self.head = newNode
 self.last.next = newNode
 self.last = newNode
 temp = self.head
 p = self.head.next
 while p != None :
 if p.cargo > newNode.cargo:
 temp = temp.next
 p = p.next
 break
 newNode.next = temp.next
 temp.next = newNode
 # Removes and returns the next item from the queue, which is the item with 
 # the highest priority. If two or more items have the same priority, those 
 # items are removed in FIFO order. An item cannot be dequeued from an
 # empty queue. The linked list is searched to find the entry with the 
 # highest priority.
 def dequeue( self ) :
 cargo = self.head.cargo
 self.head = self.head.next
 self.length = self.length - 1
 if self.length == 0:
 self.last = None
 return cargo

• Phrancis: It's unrealistic to do a thorough code review if your code is due tomorrow. We're glad you found the site, and I'm sure you'll get some good answers, but for now I would suggest to test for correctness so that you don't turn in some incorrect code for your assignment.
• Lindsey: @sᴉɔuɐɹɥԀ Sorry I meant to put Thursday, my computer autotyped.
• Phrancis: OK. I made a few edits to make your question a bit better formatted. Hope you get some great answers!

Answer 1

A Priority Queue is synonymous with a heap so I'm assuming that the implication in this assignment was to implement a maximally efficient data structure.

The optimal implementation is spelled out in Wikipedia: http://en.m.wikipedia.org/wiki/Heap_(data_structure)

If it were me, I would use a tree instead of a list. There is no Node class. A Heap object has two nullable Heap children, and every operation is recursive in nature. Length, for example, would just return each child's length or 1 if no children existed. The other operations would take advantage of the tree's structure in order to return results in logarithmic time.

• Gareth Rees: Python has a built-in heap implementation (see the heapq module) so it would make sense to use that and not write your own. Also, Python doesn't have "nullable" types as such — perhaps you are thinking of C#?
• Scott Lobdell: Yes, you are correct on both fronts. I meant "nullable" in the sense that the reference will either be a Heap or None
• Lindsey: my assignment specifically says to use linked lists!!!!

Answer 2

You are not using a priority variable for prioritizing the order of elements in the queue. Also, there are subtle issues in the code which can be fixed. Please find the below code:-

# Python code to implement Priority Queue using Linked List
# Node class 
class Node:
 def __init__(self, item, priority):
 self.item = item
 self.next = None
 self.priority = priority
class PriorityQueue:
 def __init__(self):
 self.front = self.rear = None
 # Returns a boolean value indicating whether the queue is empty
 def isEmpty(self):
 return self.front == None
 # Adds the given item to the queue by inserting it in the proper 
 # position based on the given priority. The new node is appended to 
 # the end of the linked list
 def enqueue(self, item, priority):
 newNode = Node(item, priority)
 if not self.rear:
 self.front = self.rear = newNode
 return
 if self.front.priority < newNode.priority:
 newNode.next = self.front
 self.front = newNode
 return
 previous = None
 current = self.front
 while(current and newNode.priority < current.priority):
 previous = current
 current = current.next
 if current:
 previous.next = newNode
 newNode.next = current
 else:
 self.rear.next = newNode
 self.rear = newNode
 # Removes and returns the next item from the queue, which is the 
 # item with the highest priority. If two or more items have the 
 # same priority, those items are removed in FIFO order. An item 
 # cannot be dequeued from an empty queue. 
 def dequeue(self):
 if self.isEmpty():
 print('Queue is empty')
 return
 temp = self.front
 self.front = self.front.next
 if self.front == None:
 self.rear = None
 return temp.item

• Sᴀᴍ Onᴇᴌᴀ: Welcome to Code Review and thanks for offering an answer! Would you please explicitly state above the code what you changed from the OP’s code and what justification makes it better? Please read Why are alternative solutions not welcome?
• Mast: You have presented an alternative solution, but barely reviewed the code. Please explain your reasoning (how your solution works and why it is better than the original) so that the author and other readers can learn from your thought process.