A-040● Day 402026.06.15mediumleetcode #2neetcode150

Add Two Numbers

#linked-list#math#recursion

Problem

· problem

P.040

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.

constraints

· 1 ≤ number of nodes in each list ≤ 100
· 0 ≤ Node.val ≤ 9
· No leading zeros except for the number 0 itself

// paraphrased summary — see source for full text

examples

example 1input → output

[2,4,3]
[5,6,4]

[7,0,8]

example 2input → output

[0]
[0]

[0]

example 3input → output

[9,9,9,9,9,9,9]
[9,9,9,9]

[8,9,9,9,0,0,0,1]

Pre-solve

· pre-solve

● 1list shown→● 2select→● 3reveal

☐
Can the two lists have different lengths?
☐
What happens if the final addition produces a carry? (e.g., 999 + 1 = 1000)
☐
Can we modify the input lists?
☐
Can a node's value exceed 9? Could we have a node with value 12?
☐
Why do we return dummy.next instead of dummy itself?
☐
Would it be more efficient to reverse both lists, add them, then reverse the result?
☐
Do we need to handle negative numbers?
☐
Doesn't the dummy node stay at the front of the result?

check the items you would ask, then press confirm

// session-only state — refresh resets (repeatable practice)

Logic Structure

· logic

● 1slots shown→● 2pick per slot→● 3reveal

// pick one code line per slot to assemble the algorithm flow. no typing — just the logic skeleton.

step 1· Create dummy node

○

dummy = ListNode()

○

cur = ListNode()

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dummy = l1

step 2· Initialize carry

○

carry = 0

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carry = 1

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carry = l1.val + l2.val

step 3· Loop while nodes or carry exist

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while l1 or l2 or carry:

○

while l1 or l2:

○

while l1 and l2:

step 4· Extract node values│ nested

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v1 = l1.val if l1 else 0

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v1 = l1.val if l1 else None

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v1 = l1.next.val if l1 else 0

step 5· Sum current digits and carry│ nested

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val = v1 + v2 + carry

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val = v1 + v2

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val = (v1 + v2) * carry

step 6· Extract carry via integer division│ nested

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carry = val // 10

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carry = val % 10

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carry = val / 10

step 7· Create and append new node│ nested

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cur.next = ListNode(val)

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cur = ListNode(val)

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cur.next = ListNode(val // 10)

pick one option per slot

// format: slot — recursive / DP patterns use ordering / state-first formats. ADR-08 follow-up.

Solve · Trace

· solve

solution.py

# 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

            # new digit

            val = v1 + v2 + carry

            carry = val // 10

            val = val % 10

            cur.next = ListNode(val)

            # update ptrs

            cur = cur.next

            l1 = l1.next if l1 else None

            l2 = l2.next if l2 else None

        return dummy.next

mental dry-run cases

// walk each case in your head; expand the worked example below if stuck.

case 1

[2,4,3]
[5,6,4]

→

[7,0,8]

case 2

[0]
[0]

→

[0]

case 3

[9,9,9,9,9,9,9]
[9,9,9,9]

→

[8,9,9,9,0,0,0,1]

// UI does not walk-through — you do the dry-run mentally. Expand the worked example if stuck.