# Program to find numbers represented as linked lists in Python

Suppose we have two singly linked list L1 and L2, each representing a number with least significant digits first, we have to find the summed linked list.

So, if the input is like L1 = [5,6,4] L2 = [2,4,8], then the output will be [7, 0, 3, 1, ]

To solve this, we will follow these steps:

• carry := 0

• res := a new node with value 0

• curr := res

• while L1 is not empty or L2 is not empty or carry is non-zero, do

• l0_val := value of L1 if L1 is not empty otherwise 0

• l1_val := value of L2 if L2 is not empty otherwise 0

• sum_ := l0_val + l1_val

• carry := quotient of (sum_ + carry) / 10

• add_val := remainder of (sum_ + carry) / 10

• curr.next := a new node with value add_val

• curr := next of curr

• L1 := next of L1 if L1 is not empty otherwise null

• L2 := next of L2 if L2 is not empty otherwise null

• next of curr := null

• return next of res

Let us see the following implementation to get better understanding:

## Example

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):
carry = 0
res = ListNode(0)
curr = res

while L1 or L2 or carry:
l0_val = L1.val if L1 else 0
l1_val = L2.val if L2 else 0
sum_ = l0_val + l1_val
carry, add_val = divmod(sum_ + carry, 10)

curr = curr.next

L1 = L1.next if L1 else None
L2 = L2.next if L2 else None

curr.next = None

return res.next

ob = Solution()
L1 = make_list([5,6,4])
L2 = make_list([2,4,8])
print_list(ob.solve(L1, L2))

## Input

[5,6,4], [2,4,8]

## Output

[7, 0, 3, 1, ]