One confusion when creating linked list with list

November 6, 2022

I am using below function to creating a linked list, head1, with the list, ll. It works very well. But I just confuse why using the commented part, the linked list cannot be created successfully, compared with its above three lines? What is the detailed reason?

def head_list_to_linkedlist(ll):
    head1 = ListNode(ll[0])
    for val in ll[1:]:
        node = ListNode(val)
        node.next = head1
        head1 = node
        #ListNode(val).next = head1
        #head1 = ListNode(val)
    return head1

>Solution :

The commented code creates two nodes. The first one is lost because it is not captured with a name, while the second one has nothing to do with the first, so in the end it is no different from only doing head1 = ListNode(val)

If your purpose is to somehow shorten the code a bit, then take these actions:

Make sure that the ListNode constructor can take an optional second argument which will be the value of the next attribute of the constructed node:
```
class ListNode:
    def __init__(self, val nxt=None):
        self.val = val
        self.next = nxt
```

Then, use that argument to prefix values to the linked list. To make sure you get them in the right order, iterate the ll list in reverse:

def head_linkedlist_to_list(ll):
    head = None
    for val in reversed(ll):
        head = ListNode(val, head)  # the old head becomes the next reference
    return head

This has as additional advantage that it will also work when ll is empty.