LeetCode: Add Two Numbers Python 3 iterative solution
Tags: algorithms, problemsolving
Published: 2020-03-15 15:50:06

---
title: LeetCode: Add Two Numbers Python 3 iterative solution
published: true
description: Solution for the Add Two Numbers problem from LeetCode.
tags: algorithms, problemsolving
---

Hello,

Here’s my solution for the [add two numbers](https://leetcode.com/problems/add-two-numbers/) problem on LeetCode.

*Example:*

```
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.
```

When doing problems on websites such as LeetCode, I like to code in my own IDE, that's why you’ll see some additional helper functions besides the solution code.

```python
# Definition for singly-linked list.
class ListNode(object):
    def __init__(self, x):
        self.val = x
        self.next = None

    @staticmethod
    def make_list(arr):
        if len(arr) > 0:
            root =  ListNode(arr[0])
            root.next = ListNode.make_list(arr[1:])
            return root
        else:
            return None

    @staticmethod
    def print_list(l):
        if not l:
            return
        print(l.val, end=" ")
        ListNode.print_list(l.next)

class Solution(object):
    def _add_to_list(self, root, val):
        while True:
            if root.next is None:
                root.next = ListNode(val)
                break
            else:
                root = root.next

    def _compute(self, result, base=10):
        carry = result // base
        rest = result % base
        return carry, rest

    def addTwoNumbers(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """
        root_node = None

        term_1 = l1
        term_2 = l2
        saved_carry = 0
        while True:
            # end of expression ; add cary if any
            if term_1 is None and term_2 is None:
                if saved_carry > 0:
                    self._add_to_list(root_node, saved_carry)
                break

            # We haven't reached the end, we assume that whatever term is left is 0
            if term_1 is None:
                term_1 = ListNode(0)
            if term_2 is None:
                term_2 = ListNode(0)

            # add the two terms
            term_sum_carry, term_sum_no_carry = self._compute(term_1.val + term_2.val)

            # add sum to current carry
            next_carry, total_with_carry = self._compute(term_sum_no_carry + saved_carry)

            # save carry
            saved_carry = term_sum_carry + next_carry

            # save total
            if not root_node:
                root_node = ListNode(total_with_carry)
            else:
                self._add_to_list(root_node, total_with_carry)

            # move on
            term_1 = term_1.next
            term_2 = term_2.next

        return root_node

if __name__ == '__main__':
    l1 = ListNode.make_list([9, 9])
    l2 = ListNode.make_list([1])
    solution = Solution()
    result = solution.addTwoNumbers(l1, l2)
    ListNode.print_list(result)
    print("Done")
```

My solution isn’t that great because of the *_add_to_list* function which iterates the root_node to append the next value. I have no idea how to optimize that right now.

Thanks for reading!