Suppose we have two linked lists l1 and l2, we have to return one linked list by interleaving elements of these two lists starting with l1. If there are any leftover nodes in a linked list, they should be appended to the list.
So, if the input is like l1 = [5,4,6,3,4,7] l2 = [8,6,9], then the output will be [5,8,4,6,6,9,3,4,7]
To solve this, we will follow these steps −
ans := l1
while l2 is not null, do
if ans is not null, then
if next of ans is not null, then
newnode := a new list node with same value of l2
next of newnode := next of ans
next of ans := newnode
ans := next of newnode
l2 := next of l2
otherwise,
next of ans := l2
come out from the loop
otherwise,
return l2
return l1
Let us see the following implementation to get better understanding −
Example
Source Code (Python):
class ListNode:
def __init__(self, data, next = None):
self.val = data
self.next = next
def make_list(elements):
head = ListNode(elements[0])
for element in elements[1:]:
ptr = head
while ptr.next:
ptr = ptr.next
ptr.next = ListNode(element)
return head
def print_list(head):
ptr = head
print('[', end = "")
while ptr:
print(ptr.val, end = ", ")
ptr = ptr.next
print(']')
class Solution:
def solve(self, l1, l2):
ans = l1
while l2:
if ans:
if ans.next != None:
newnode = ListNode(l2.val, None)
newnode.next = ans.next
ans.next = newnode
ans = newnode.next
l2 = l2.next
else:
ans.next = l2
break
else:
return l2
return l1
ob = Solution()
l1 = make_list([5,4,6,3,4,7])
l2 = make_list([8,6,9])
res = ob.solve(l1,l2)
print_list(res)Input
[5,4,6,3,4,7],[8,6,9]
Output
[5, 8, 4, 6, 6, 9, 3, 4, 7, ]