README.md

2. Add Two Numbers

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 contain a single digit. Add the two numbers and return it as a linked list.

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

Example:

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

Solutions (Python)

1. Solution

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

class Solution:
    def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
        l3 = ListNode(0)
        curr = l3
        carry = 0

        while l1 or l2:
            curr.next = ListNode(carry)
            curr = curr.next
            curr.val += l1.val if l1 else 0
            curr.val += l2.val if l2 else 0
            carry = curr.val // 10
            curr.val %= 10

            l1 = l1.next if l1 else None
            l2 = l2.next if l2 else None

        if carry:
            curr.next = ListNode(carry)

        return l3.next

Solution (Ruby)

1. Solution

# Definition for singly-linked list.
# class ListNode
#     attr_accessor :val, :next
#     def initialize(val = 0, _next = nil)
#         @val = val
#         @next = _next
#     end
# end
# @param {ListNode} l1
# @param {ListNode} l2
# @return {ListNode}
def add_two_numbers(l1, l2)
    l3 = ListNode.new
    curr = l3
    carry = 0

    while l1 or l2
        curr.next = ListNode.new(carry)
        curr = curr.next
        curr.val += l1.val if l1
        curr.val += l2.val if l2
        carry = curr.val / 10
        curr.val %= 10

        l1 = l1.next if l1
        l2 = l2.next if l2
    end

    curr.next = ListNode.new(carry) if carry == 1

    return l3.next
end