LC2. Add Two Numbers¶

Problem Description¶

LeetCode Problem 2: 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.

Clarification¶

two linked list have different number of nodes
Modify the existing node?
Reverse order

Assumption¶

Solution¶

Approach - Elementary Math¶

Follow the elementary math on sum, begin by summing the least-significant digits, which are the heads of l1 and l2. Summing two digits may overflow and needs additional variable carry to store overflow value.

Python

class Solution:
    def addTwoNumbers(self, l1: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]:
        dummy = ListNode(-1, None)

        p1 = l1
        p2 = l2
        p = dummy
        carry = 0
        while p1 or p2 or carry > 0:
            if p1 is None:
                val1 = 0
            else:
                val1 = p1.val
                p1 = p1.next

            if p2 is None:
                val2 = 0
            else:
                val2 = p2.val
                p2 = p2.next

            sum_node = val1 + val2 + carry
            carry = sum_node // 10
            p.next = ListNode(sum_node % 10, None)

            p = p.next

        return dummy.next

Complexity Analysis¶

Time complexity: \(O(n)\)
Need to go through \(max(n_1, n_2)\) nodes, where \(n_1\) is the number of nodes of l1 and \(n_2\) is the number of nodes of l2.
Space complexity: \(O(n)\)
Need to create a linked list to store the sum, which contains at most \(max(n1, n2) + 1\) nodes.

Test¶

one list is longer than the other
one list is an empty list
the sum could have an extra carry of one at the end