How do you solve Add Two Numbers II on LeetCode?

Add Two Numbers II (LeetCode #445) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using stacks allows us to reverse the order of digits easily.

What is the brute force solution for Add Two Numbers II?

The brute force approach for Add Two Numbers II is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves converting both linked lists into their respective integer values, adding them together, and then converting the result back into a linked list. This is straightforward but not efficient for larger inputs. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Add Two Numbers II?

Add Two Numbers II uses the following concepts: Linked List, Math, Stack. The recommended patterns to study are: Stack, Two Pointers.

What are the interview tips for Add Two Numbers II?

Explain your thought process clearly before coding. Consider edge cases, like different lengths of linked lists. Practice writing clean and efficient code.

What are common mistakes when solving Add Two Numbers II?

Forgetting to handle the carry at the end of addition. Not properly managing the linked list pointers.

#445

Add Two Numbers II

Medium

Linked ListMathStackStackTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses stacks to reverse the order of the digits and then adds them together. This avoids the need for converting the entire linked list into a number, making it more efficient and straightforward.

⚙️

Algorithm

3 steps

1Step 1: Use two stacks to store the digits of both linked lists.
2Step 2: Pop the digits from the stacks and add them together, keeping track of the carry.
3Step 3: Construct the resulting linked list from the sum, ensuring to handle any remaining carry.

solution.py27 lines

1# Full working Python code
2class ListNode:
3    def __init__(self, val=0, next=None):
4        self.val = val
5        self.next = next
6
7def addTwoNumbers(l1, l2):
8    stack1, stack2 = [], []
9    while l1:
10        stack1.append(l1.val)
11        l1 = l1.next
12    while l2:
13        stack2.append(l2.val)
14        l2 = l2.next
15    carry = 0
16    dummy = ListNode()
17    while stack1 or stack2 or carry:
18        sum = carry
19        if stack1:
20            sum += stack1.pop()
21        if stack2:
22            sum += stack2.pop()
23        carry = sum // 10
24        new_node = ListNode(sum % 10)
25        new_node.next = dummy.next
26        dummy.next = new_node
27    return dummy.next

ℹ

Complexity note: The time complexity is O(n) because we traverse each linked list once. The space complexity is O(n) due to the stacks used to store the digits.

1Using stacks allows us to reverse the order of digits easily.
2Handling carry is crucial when adding the digits.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.