a = b = 0 in Python

Question

I was practicing some algorithms, and I once saw this code

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

class Solution:
    def addTwoNumbers(self, l1: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]:
        
        q = 0
        head = new_LN = ListNode(0)
         
        while l1 or l2:
            l1_val = l1.val if l1 else 0
            l2_val = l2.val if l2 else 0
            q, r = divmod(q + l1_val + l2_val , 10)
            new_LN.next = ListNode(r)
            new_LN = new_LN.next
            l1 = l1.next if l1 else None
            l2 = l2.next if l2 else None
        if q:
            new_LN.next = ListNode(q)
            
        return head.next

    # print(new_LN) # ListNode{val: 8, next: None}
    # print(head) # ListNode{val: 0, next: ListNode{val: 7, next: ListNode{val: 0, next: ListNode{val: 8, next: None}}}}

In an example, the results of new_LN and head are stated above.

I cannot understand why head and new_LN are different though they were declared on the same line.

Can anybody explain this one for me?

Ohad Sharet Ohad Sharet · Accepted Answer · 2022-07-28T16:24:42

what you need to understand is that both head and new_LN are pointers to an object and not the object itself, so at the beginning of the code when the line head = new_LN = ListNode(0) is called a new ListNode object is created and both head and new_LN are pointing to it. but when the line new_LN = new_LN.next is called, the pointer new_LN is set to point to another object while head still points to the original node, so that's why at the end of the code head and new_LN are not equal

hope I could clarify that for you if you have any questions feel free to ask me in the comments, and if my comment helped you marking it as the answer or upvoting will go a long way :)

a = b = 0 in Python

1 Answers