# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
# multiple assignment, so we can do the return at the end
start = cur = ListNode(0)
while l1 and l2:
if l1.val >= l2.val:
cur.next = l2
l2 = l2.next
else:
cur.next = l1
l1 = l1.next
cur = cur.next
# either l1 or l2 is done
if l1:
cur.next = l1
else:
cur.next = l2
# done
return start.next
# Other solutions I found interesting:
class Solution:
# A recursive solution
def mergeTwoLists(self, a, b):
if a and b:
if a.val > b.val:
a, b = b, a
a.next = self.mergeTwoLists(a.next, b)
return a or b
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