LeetCode Problem

Problem Description

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.

Examples

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

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

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.

Follow-up

Approach to Solve

Traverse both lists simultaneously, adding corresponding digits and keeping track of the carry. Create a new linked list to store the result.

Code Implementation

class Solution:
    def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
        dummy = ListNode(0)
        current = dummy
        carry = 0
        while l1 or l2 or carry:
            val1 = l1.val if l1 else 0
            val2 = l2.val if l2 else 0
            total = val1 + val2 + carry
            carry = total // 10
            current.next = ListNode(total % 10)
            current = current.next
            l1 = l1.next if l1 else None
            l2 = l2.next if l2 else None
        return dummy.next

class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode dummy = new ListNode(0);
        ListNode current = dummy;
        int carry = 0;
        while (l1 != null || l2 != null || carry != 0) {
            int val1 = (l1 != null) ? l1.val : 0;
            int val2 = (l2 != null) ? l2.val : 0;
            int total = val1 + val2 + carry;
            carry = total / 10;
            current.next = new ListNode(total % 10);
            current = current.next;
            l1 = (l1 != null) ? l1.next : null;
            l2 = (l2 != null) ? l2.next : null;
        }
        return dummy.next;
    }
}

class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        ListNode* dummy = new ListNode(0);
        ListNode* current = dummy;
        int carry = 0;
        while (l1 || l2 || carry) {
            int val1 = l1 ? l1->val : 0;
            int val2 = l2 ? l2->val : 0;
            int total = val1 + val2 + carry;
            carry = total / 10;
            current->next = new ListNode(total % 10);
            current = current->next;
            l1 = l1 ? l1->next : nullptr;
            l2 = l2 ? l2->next : nullptr;
        }
        return dummy->next;
    }
};

Explanation

We iterate through both linked lists simultaneously, adding the corresponding digits and any carry from the previous addition. We create a new linked list to store the result. The time and space complexity are both O(max(n, m)), where n and m are the lengths of the input linked lists.

Complexity

Time Complexity: O(max(n, m))
Space Complexity: O(max(n, m))

LeetSolved

Add Two Numbers