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List Does Not Change In Attempt To Solve "Delete Node In Doubly Linked List"

- 1 answer

I am working on the GeeksForGeeks problem Delete node in Doubly Linked List:

Given a doubly linked list and a position. The task is to delete a node from given position in a doubly linked list.

Your Task:

The task is to complete the function deleteNode() which should delete the node at given position and return the head of the linkedlist.

My code:

def deleteNode(self,head, x):
    # Code here
    temp=head        
    count_of_nodes=0
    prev_of_delete_node=None
    next_of_delete_node=None
    while temp != head:
        count_of_nodes+=1
        if count_of_nodes==x:
            prev_of_delete_node=temp.prev
            next_of_delete_node=temp.next
            #print(y.data,z.data)
            prev_of_delete_node.next=next_of_delete_node
            next_of_delete_node.prev=prev_of_delete_node
            break
        
        temp=temp.next
    
    if x==1:
        head=next_of_delete_node

There is no effect on the doubly LinkedList after executing above code. Why is this?

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Answer

Some issues:

  • The while condition is wrong: it is false immediately, so the loop will not execute.

  • The value for prev_of_delete_node could be None when you dereference it with prev_of_delete_node.next. So guard that operation. Same for next_of_delete_node.

  • The function doesn't return anything, but it should return the head of the list after the deletion

Correction:

def deleteNode(self,head, x): 
    temp=head
    count_of_nodes=0
    prev_of_delete_node=None
    next_of_delete_node=None
    while temp:  # Corrected loop condition
        count_of_nodes+=1
        if count_of_nodes==x:
            prev_of_delete_node=temp.prev
            next_of_delete_node=temp.next
            if prev_of_delete_node:  # Guard
                prev_of_delete_node.next=next_of_delete_node
            if next_of_delete_node:  # Guard
                next_of_delete_node.prev=prev_of_delete_node
            break
        
        temp=temp.next
    # Should return:
    if x==1:
        return next_of_delete_node
    return head 
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source: stackoverflow.com
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