Add Two Numbers

#datastructures #leetcode #interview

2. Add Two Numbers

Medium

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.

Python Solution.


class Solution:
    def addTwoNumbers(self, l1: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]:

        head = None
    # Reference of head which is null at this poin
        temp = None
    # Carry
        carry = 0
    # Loop for the two lists
        while l1 is not None or l2 is not None:

        # At the start of each iteration, we should add carry from the last iteration
            sum_value = carry
        # Since the lengths of the lists may be unequal, we are checking if the
        # current node is null for one of the lists
            if l1 is not None:
                sum_value += l1.val
                l1 = l1.next
            if l2 is not None:
                sum_value += l2.val
                l2 = l2.next
        # At   this point, we will add the total sum_value % 10 to the new node
        # in the resultant list
            node = ListNode(sum_value % 10)
        # Carry to be added in the next iteration
            carry = sum_value // 10
        # If this is the first node or head
            if temp is None:
                temp = head = node
        # for any other node
            else:
                temp.next = node
                temp = temp.next
    # After the last iteration, we will check if there is carry left
    # If it's left then we will create a new node and add it
        if carry > 0:
            temp.next = ListNode(carry)
        return head