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 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.

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.

Approach 1:

Recursive Solution

We go on adding both of the node values and the carry in a recursive manner , once we find any of the linked list is exhausted then the value assumed as 0 . Also we link each of the node to its previous node at the end of the recursion

class Solution(object):
    def addTwoNumbers(self, l1, l2,carry=0):
        #print(l1.val,l2.val,bool(l1),bool(l2))
        val1,val2=0,0
        
        val1=l1.val if l1 else 0
        val2=l2.val if l2 else 0
        carry,quotient=divmod(val1+val2+carry,10)
        node=ListNode(quotient)
        l1= l1.next if l1 else l1
        l2= l2.next if l2 else l2     
        if l1 or l2 or carry:
            node.next=self.addTwoNumbers(l1,l2,carry)
        return node

Time: O(max(m,n)) m,n -> size of first and second linked list
Space: O(max(m,n))

Approach 2:

Same can be solved using the Iterative solution.

class Solution(object):
    def addTwoNumbers(self, l1, l2):
        carry=0
        temp=ListNode(0)
        node_=temp
        while l1 or l2:
            if l1 ==None:
                one=0
            else:
                one=l1.val
                
            if l2==None:
                two=0
            else:
                two=l2.val
            inter=ListNode((one+two+carry)%10)    
            temp.next=inter
            temp=inter
            carry=(one+two+carry)//10
            if l1:
                l1=l1.next
            if l2:
                l2=l2.next
        if carry!=0:
            temp.next=ListNode(carry)
        
        return node_.next

Time: O(max(m,n)) m,n -> size of first and second linked list
Space: O(max(m,n))