Why is it necessary to take a new variable "head" in for loop?

Question

class Node:
    def __init__(self, data):
        self.data = data
        self.ref = None

def Print_LL(linkedlist):
    if linkedlist is None:
        print("LinkedList is empty!!")
    else:
        while linkedlist is not None:
            print(linkedlist.data)
            linkedlist = linkedlist.ref

def arr_to_LL(arr, n):
    linkedlist = None

    for i in range(0,n):
        new_node = Node(arr[i])

        if linkedlist is None:
            linkedlist = new_node
        else:
            head = linkedlist
            while head.ref is not None:
                head = head.ref
            head.ref = new_node
    return linkedlist

why to take "head" variable? why cant we continue with linkedlist

Asked By: sai Rohit

||

Source

Answer 1

linkedlist is a reference to the head of the list. head, despite the name, is just a reference to the end of the list, or more specifically to the node to which you’ll attach a new node. A better definition of arr_to_LL might be

def arr_to_LL(arr):
    head = None

    for v in arr:
        new_node = Node(v)

        if head is None:
            head = new_node
        else:
            curr.ref = new_node

        curr = new_node
           
    return head

This for loop has two invariants:

head always refers to the front of the (possibly empty) linked list.
curr (once defined) always refers to the end of the linked list.

As you iterate over the array, you simply append a new node to the end, then update the end.

If you use the common convention of an empty linked list represented by a dummy node, it becomes even simpler, as now you don’t need a special case for head is None.

def arr_to_LL(arr):
    # We'll store the length in the dummy node, because why not?
    head = curr = Node(0)

    for v in arr:
        curr.ref = Node(v)
        curr = new_node
        head.data += 1
           
    return head

Answered By: chepner

Why is it necessary to take a new variable "head" in for loop?

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