linked_list_add_two_numbers.py

'''
You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order, and each of their nodes contains a single digit. Add the two numbers and return the sum as a linked list.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Example 1:
Input: l1 = [2,4,3], l2 = [5,6,4]
Output: [7,0,8]
Explanation: 342 + 465 = 807.

Example 2:
Input: l1 = [0], l2 = [0]
Output: [0]

Example 3:
Input: l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]
Output: [8,9,9,9,0,0,0,1]
 

Constraints:

The number of nodes in each linked list is in the range [1, 100].
0 <= Node.val <= 9
It is guaranteed that the list represents a number that does not have leading zeros.

Complexity
Time complexity : O(max(l1,l2)).
Space complexity : O(n).

'''

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def addTwoNumbers(self, l1: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]:
        carry = 0
        head = ListNode('*');
        header_pointer = head;
        while(l1 != None or l2 != None):
            temp = carry
            if(l1 == None):
                temp += l2.val
                l2 = l2.next
            elif(l2 == None):
                temp += l1.val
                l1 = l1.next
            else:
                temp += l1.val+l2.val
                l1 = l1.next
                l2 = l2.next
            head.next = ListNode(temp % 10)
            carry = temp // 10
            head = head.next
        if(carry != 0):
            head.next = ListNode(carry)
            head = head.next
        return header_pointer.next