First I didn't understand the problem I think. In the sense that I accepted the inputs as two lists and returned a list - Python list. Then I read it and realized they want the operation in Linked lists.
i took some time to recall how to move along a list and finally figured it out.
The other thing that gave me a minor ankle break was when the test case was two empty lists.
Handled that as well eventually.
Here's the cumulative code -
class Solution:
def mergeTwoLists(self, list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[ListNode]:
c = []
a =[]
b=[]
i = list1
j = list2
while i:
a.append(i.val)
i = i.next
while j:
b.append(j.val)
j = j.next
i, j = 0 , 0
while i < len(a) and j < len(b):
if a[i] < b[j]:
c.append(a[i])
i = i + 1
else:
c.append(b[j])
j = j+1
if j < len(b):
c.extend(b[j:])
elif i < len(a):
c.extend(a[i:])
if len(c) > 0:
listNode3 = ListNode(c[0],None)
head = listNode3
for i in c[1:]:
temp_node = ListNode(i,None)
listNode3.next = temp_node
listNode3 = listNode3.next
else:
head = None
return head
And boom!
By minorly changing the above code to not create lists out of ListNodes initially, a speedup of 2 seconds can be achieved -
class Solution:
def mergeTwoLists(self, list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[ListNode]:
c = []
a =[]
b=[]
i = list1
j = list2
while i and j:
if i.val < j.val:
c.append(i.val)
i = i.next
else:
c.append(j.val)
j = j.next
if j:
while j:
c.append(j.val)
j = j.next
elif i:
while i :
c.append(i.val)
i = i.next
if len(c) > 0:
listNode3 = ListNode(c[0],None)
head = listNode3
for i in c[1:]:
temp_node = ListNode(i,None)
listNode3.next = temp_node
listNode3 = listNode3.next
else:
head = None
return head
