# Program to find union of two given linked lists in Python

Suppose we have two sorted linked lists L1 and L2, we have to return a new sorted linked list that is the union of the two given lists.

So, if the input is like L1 = [10,20,30,40,50,60,70] L2 = [10,30,50,80,90], then the output will be [10, 20, 30, 40, 50, 60, 70, 80, 90, ]

To solve this, we will follow these steps −

• Define a function solve() . This will take L1, L2
• if L1 is empty, then
• return L2
• if L2 is empty, then
• return L1
• if value of L1 < value of L2, then
• res := L1
• next of res := solve(next of L1, L2)
• otherwise when value of L2 < value of L1, then
• res := L2
• next of res := solve(next of L2, L1)
• otherwise,
• res := L1
• next of res := solve(next of L1, next of L2)
• return res

Let us see the following implementation to get better understanding −

## Example

Live Demo

class ListNode:
def __init__(self, data, next = None):
self.val = data
self.next = next
def make_list(elements):
for element in elements[1:]:
while ptr.next:
ptr = ptr.next
ptr.next = ListNode(element)
print('[', end = "")
while ptr:
print(ptr.val, end = ", ")
ptr = ptr.next
print(']')
class Solution:
def solve(self, L1, L2):
if not L1:
return L2
if not L2:
return L1
if L1.val < L2.val:
res = L1
res.next = self.solve(L1.next, L2)
elif L2.val < L1.val:
res = L2
res.next = self.solve(L2.next, L1)
else:
res = L1
res.next = self.solve(L1.next, L2.next)
return res
ob = Solution()
L1 = make_list([10,20,30,40,50,60,70])
L2 = make_list([10,30,50,80,90])
print_list(ob.solve(L1, L2))

## Input

[10,20,30,40,50,60,70], [10,30,50,80,90]

## Output

[10, 20, 30, 40, 50, 60, 70, 80, 90, ]