2. Add Two Numbers

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LeetCode

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Problem

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 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.

Code

class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode dummy = new ListNode(-1);
        ListNode curr = dummy;

        int carry = 0;
        while(l1 != null || l2 != null){
            int x = l1 == null ? 0 : l1.val;
            int y = l2 == null ? 0 : l2.val;

            int sum = x + y + carry;
            curr.next = new ListNode(sum % 10);
            carry = sum / 10;
            curr = curr.next;

            if(l1 != null) l1 = l1.next;
            if(l2 != null) l2 = l2.next;
        }

        if(carry != 0){
            curr.next = new ListNode(carry);
        }
        return dummy.next;
    }
}

class Solution:
    def addTwoNumbers(
            self, l1: Optional[ListNode], l2: Optional[ListNode]
    ) -> Optional[ListNode]:
        dummy = ListNode(0)
        curr = dummy
        carry = 0
        while l1 or l2:
            num1 = l1.val if l1 else 0
            num2 = l2.val if l2 else 0
            sum = num1 + num2 + carry
            curr.next = ListNode(sum % 10)
            carry = sum // 10  # performs floor division, 7 // 2 = 3
            l1 = l1.next if l1 else None
            l2 = l2.next if l2 else None
            curr = curr.next
        if carry:
            curr.next = ListNode(carry)
        return dummy.next