[Python] Why do I need to assign to `output.next` in this solution for Merge Sorted Linked...

The correct solution below for the Merge Two Sorted Linked Lists Leetcode problem fails when you remove the .next in lines 12, 15, 24:

output.next = list1 becomes output = list1

output.next = list2 becomes output = list2

return head.next becomes return head

Correct solution


# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def mergeTwoLists(self, list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[ListNode]:
head = output = ListNode()

while list1 and list2:
if list1.val < list2.val:
output.next = list1
list1 = list1.next
else:
output.next = list2
list2 = list2.next
output = output.next

if list1 and output:
output.next = list1
elif list2 and output:
output.next = list2

return head.next


I expected it to work exactly the same. When I print output within the loop, it prints the remaining tail of the current list that's lesser than.

Continue reading...