CODEWITHSUNDEEP
pythonpython-3.xlinked-list
4daJKong
4daJKong

Reputation: 2045

How to create a linked list by python in this class

I see a linked list class definition in Leetcode:

class ListNode:
    def __init__(self, val=0, next=None):          
        self.val = val
        self.next = None

I am confused how to use it, there are two methods I want to figure out:

Firstly: if I want to create a linked list 1->3->5 step by step, how? My idea is that:

p1 = ListNode(1)
p2 = ListNode(3)
p3 = ListNode(5)

p2 = p1.next
p3 = p1.next.next
print(p1.val)  #however it shows p1 doesn't have next attr...

Secondly, if I have a linked list, like that:

l1 = ListNode([1, 2, 4, 6])

How to get every value step by step? I tried like:

print(l3.next) 
print(l3.next.next)#maybe just some false idea...

Upvotes: 2

Views: 406

Answers (1)

Konrad Rudolph
Konrad Rudolph

Reputation: 545518

The constructor is wrong: it sets self.next to None rather than to next. Let’s fix that:

class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

Now we can create a list of three nodes. Since this is a singly-linked forward list, we need start at the back:

p3 = ListNode(3)
p2 = ListNode(2, p3)
p1 = ListNode(1, p2)

print(p1.next.next.val) # 3

Or, to generate a linked list from a list:

def from_list(lst) -> ListNode:
    node = None
    for item in lst[::-1]:
        node = ListNode(item, node)

    return node

x = from_list([1, 2, 3, 4])
print(x.next.next.next.val) # 4

Upvotes: 2

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