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.

Java:

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        if(l1 == null)
            return l2;
        if(l2 == null)
            return l1;
        ListNode head = new ListNode(0);
        ListNode p = head;
         
        int temp = 0;
        while(l1 != null || l2 != null || temp != 0) {
            if(l1 != null) {
                temp += l1.val;
                l1 = l1.next;
            }
            if(l2 != null) {
                temp += l2.val;
                l2 = l2.next;
            }
             
            p.next = new ListNode(temp % 10);
            p = p.next;
            temp /= 10;          
        }
        return head.next;
    }
}

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        int sum = 0;
        ListNode res = new ListNode(-1), tail = res;
        while(l1 != null || l2 != null || sum >= 10) {
            sum /= 10;
            if(l1 != null) {
                sum += l1.val;
                l1 = l1.next;
            }
            if(l2 != null) {
                sum += l2.val;
                l2 = l2.next;
            }
            tail.next = new ListNode(sum % 10);
            tail = tail.next;
        }
        return res.next;    
    }
}

Python:

# 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:
        carry = 0
        temp = l1
        num = 0
        while l2 != None or carry != 0:
            if l2 != None:
                num = l2.val
            l1.val = num + l1.val + carry
            carry = 0
            if l1.val > 9:
                l1.val -= 10
                carry = 1
            if l1.next == None:
                if l2 == None:
                    if carry != 0:
                        l1.next = ListNode(0)
                else:
                    if l2.next != None or carry != 0:
                        l1.next = ListNode(0)
            if l2 != None:
                l2 = l2.next
            l1 = l1.next
            num = 0
        return temp

    # 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:
            dummyHead = ListNode(0)
            curr, carry = dummyHead, 0
            while l1 or l2:
                sum = 0
                if l1:
                    sum += l1.val
                    l1 = l1.next
                if l2:
                    sum += l2.val
                    l2 = l2.next
                sum += carry
                carry = sum // 10
                curr.next = ListNode(sum % 10)
                curr = curr.next
            if carry > 0:
                curr.next = ListNode(1)
            return dummyHead.next;

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
        dummy = ListNode()
        cur = dummy
        
        carry = 0
        while l1 or l2 or carry:
            v1 = l1.val if l1 else 0
            v2 = l2.val if l2 else 0
            
            # compute digit
            val = v1 + v2 + carry
            carry = val // 10
            val %= 10
            cur.next = ListNode(val)
            
            # update pointers
            cur = cur.next
            l1 = l1.next if l1 else None
            l2 = l2.next if l2 else None
        
        return dummy.next

/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *     this.val = val;
 *     this.next = null;
 * }
 */
/**
 * @param {ListNode} l1
 * @param {ListNode} l2
 * @return {ListNode}
 */
var addTwoNumbers = function(l1, l2) {
    var head = new ListNode(0);
    var current = head;
    var carry = 0;
    while (l1 || l2) {
        l1 = l1 || new ListNode(0);
        l2 = l2 || new ListNode(0);
        var sum = l1.val + l2.val + carry;
        if (sum > 9) {
            carry = 1;
            sum = sum - 10;
        } else {
            carry = 0;
        }
        
        var node = new ListNode(sum);
        current.next = node;
        current = node;
        
        l1 = l1.next;
        l2 = l2.next;
    }
    
    if (carry !== 0) {
        current.next = new ListNode(carry);
    }
    return head.next
};