(Medium #2) Add Two Numbers

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:

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.

My solution

class Solution:

    def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:

        l3 = ListNode(0)

        print(l1)

     

        l1_val = ''

        while l1 != None:

            l1_val = str(l1.val) + l1_val

            l1 = l1.next

         

        l2_val = ''

        while l2 != None:

            l2_val = str(l2.val) + l2_val

            l2 = l2.next

     

        cal  = eval(l1_val + '+' + l2_val)     

        str_cal = str(cal)

        l3_last = ListNode(int(str_cal[0]))     

        for i in range(1,len(str_cal)):

            l3_next = ListNode(int(str_cal[i]))

            l3_next.next = l3_last

            l3_last = l3_next

         

        print(l3_last)

     

        return l3_last

Runtime: 124ms / Memory: 14.1MB

Other people solution

class Solution:
    def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
        head = ListNode(None)
        carry = 0
        pre = head
        while l1 or l2:
            curr = ListNode(0)
            print(l1.val if l1 else 0)
            print(l2.val if l2 else 0)
            carry, curr.val = divmod((l1.val if l1 else 0) + (l2.val if l2 else 0) + carry, 10)
            pre.next = curr
            pre = curr
            if l1: l1 = l1.next
            if l2: l2 = l2.next
        if carry:
            pre.next = ListNode(carry)

        return head.next

Runtime: 80ms / Memory: 14.1MB