CODEWITHSUNDEEP
pythonpython-3.xlinked-listsingly-linked-list
Gaurang Shah
Gaurang Shah

Reputation: 12940

how do create a linked list in python

I am trying to solve a linked list coding challenge in python. And I have given only following class to create a linked list 

# Definition for singly-linked list.
class ListNode(object):
    def __init__(self, x):
        self.val = x
        self.next = None

I can create a linked list something like this 

x = ListNode(1)
x.next = ListNode(4)
x.next.next = ListNode(5)

however, How do I create iterativly (inside a for loop)

Upvotes: 4

Views: 17384

Answers (4)

maxim-lixakov
maxim-lixakov

Reputation: 13

dummy = ListNode()  
prev = dummy
for val in arr:  
    node = ListNode(val)  
    prev.next = node  
    prev = node

dummy.next - is the link to the first element of created linked list

Upvotes: 1

DYZ
DYZ

Reputation: 57105

You need two "pointers" that memorize the head and the tail of your list. The head is initialized once. You will eventually use it to access the whole list. The tail changes every time you add another node:

data = [5, 1, 7, 96]
tail = head = ListNode(data[0])
for x in data[1:]:
    tail.next = ListNode(x) # Create and add another node
    tail = tail.next # Move the tail pointer

Upvotes: 3

Brian Rodriguez
Brian Rodriguez

Reputation: 4366

You can add a next argument to the constructor:

class ListNode(object):
    def __init__(self, x, next=None):
        self.x = x
        self.next = next

head = None
for x in [5, 1, 7, 96]:
    head = ListNode(x, next=head)

# Linked list looks like:
# (96) -> ( 7) -> ( 1) -> ( 5) -> None

Upvotes: 0

R4444
R4444

Reputation: 2114

You can do something like:

arr = [1,4,5]
for i in arr:
    x = ListNode(i)
    x = x.next

But now x will become None. As there's nothing else to keep track, you can't print the elements.

You can verify this by printing the values of x.val in between the loop:

arr = [1,4,5]
for i in arr:
    x = ListNode(i)
    print(x.val)
    x = x.next

output:

1
4
5

Upvotes: 0

Related Questions