Skip to main content

2. Add Two Numbers

https://leetcode.com/problems/add-two-numbers

Python​

Convert to list then calculate back​

from typing import Optional, List


# 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: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]:
        list1, list2 = [], []

        while l1.next is not None:
            list1.append(l1.val)
            l1 = l1.next
        list1.append(l1.val)

        while l2.next is not None:
            list2.append(l2.val)
            l2 = l2.next
        list2.append(l2.val)

        result = list(str(
            int(''.join([str(i) for i in list1[::-1]])) + \
            int(''.join([str(i) for i in list2[::-1]]))
        ))

        head = ListNode(val=result.pop(), next=None)
        cur = head
        while result:
            cur.next = ListNode(val=result.pop(), next=None)
            cur = cur.next

        return head

One-pass Linked List travsal​

class Solution:
    def addTwoNumbers(self, l1: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]:
        dummy_head = ListNode()
        cur = dummy_head
        
        cur1, cur2 = l1, l2
        carry = 0
        while cur1 or cur2:
            if not cur1:
                value = cur2.val + carry
                cur2 = cur2.next
            elif not cur2:
                value = cur1.val + carry
                cur1 = cur1.next
            else:
                value = cur1.val + cur2.val + carry
                cur1 = cur1.next
                cur2 = cur2.next

            cur.next = ListNode(val=value%10)
            carry = value // 10
            cur = cur.next
        
        if carry:
            cur.next = ListNode(val=carry)

        return dummy_head.next