Inserting In The Middle Of A Doubly Linked List- Python
I am new to stackoverflow and the Python language and have a question. I know how to implement a singly linked list in Python but am having trouble with doubly linked list, more s
Solution 1:
Me also, I'm new to Python, but I hope I can help you. So if I got your question correctly, suppose you have some (classic) linked list implementation like this:
# Simple LinkedList classclassLinkedList:
def__init__(self, data, prev=None, next=None):
self.data = data
self.prev = prev
self.next = next# Just a string representationdef__str__(self):
if self.next:
return'%s %s' % (str(self.data), self.next.__str__())
else:
return'%s' % str(self.data)
You can easily insert an element to middle, knowing the linked list's ends, by iterating simultaneously from the ends to middle. When the iterators intersect, you add your element:
# Insert to middle logicdefinsertToMiddle(data, left, right):
# IteratewhileTrue:
# Test for intersectionif left.nextis right:
node = LinkedList(data, left, right)
left.next = node
right.prev = node
returnelif left is right:
node = LinkedList(data, left.prev, right.next)
left.next = node
right.next.prev = node
return# Next iteration
left = left.next
right = right.prev
# This doesn't actually execute for a right callifnot left ornot right:
return
Below, you can see a linked list creation, and its representation, before and after the insertion:
# Sample list creation
node1 = LinkedList(1)
node2 = LinkedList(2,node1)
node3 = LinkedList(3,node2)
node4 = LinkedList(4,node3)
node1.next = node2
node2.next = node3
node3.next = node4
# Testprint node1
insertToMiddle(5, node1, node4)
print node1
insertToMiddle(6, node1, node4)
print node1
insertToMiddle(7, node1, node4)
print node1
Output:
1 2 3 4 #initial
1 2 5 3 4 #inserted 5
1 2 5 6 3 4 #inserted 6, notice it's right to middle
1 2 5 7 6 3 4 #inserted 7
Remark: If your list has an odd number of elements (say 2 3 4), inserting to middle is somehow undefined, so the above function will add immediately to middle's right (2 3 elem 4)
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