Add Two Numbers

Add Two Numbers

MediumLeetCodeLinked ListMathRecursion


This problem involves adding two numbers represented by linked lists.

You can find the original problem described on LeetCode.

Problem Statement

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.

Problem Solving

There are a few ways to solve this problem. I will explain all of them to you in the next sections. ๐Ÿ‘จโ€๐Ÿซ
Before that, below is the definition of ListNode by LeetCode, which we’re going to use on every solution.

public class ListNode {
    
    public var val: Int
    public var next: ListNode? = nil
    
    public init(_ val: Int) {
        self.val = val
    }
    
    public init(_ val: Int, _ next: ListNode?) {
        self.val = val
        self.next = next
    }
}

Iterative method

The iterative method involves traversing both linked lists, adding corresponding digits along with any carry-over from the previous addition.

func addTwoNumbers(_ l1: ListNode?, _ l2: ListNode?) -> ListNode? {
    // Dummy head to handle corner case (we return the next node)
    let dummyHead = ListNode(0)
    // This is the pointer we will use to construct the new list
    var currentNode: ListNode? = dummyHead
    // These pointers will navigate through the nodes
    var p1 = l1
    var p2 = l2
    
    // No initial carry value
    var carry = 0
    
    // Iterate until we reach the end of both lists
    while p1 != nil || p2 != nil {
        // Sum of the current values and carry
        let sum = carry + (p1?.val ?? 0) + (p2?.val ?? 0)
        
        // New carry
        carry = sum / 10
        // New node with the unit digit
        currentNode?.next = ListNode(sum % 10)
        
        // Move to the next nodes
        currentNode = currentNode?.next
        p1 = p1?.next
        p2 = p2?.next
    }
    
    // Add a new node if there's a carry left
    if carry > 0 {
        currentNode?.next = ListNode(carry)
    }
    
    // Return the head of the new list (ignore the dummy head)
    return dummyHead.next
}

Source code

Recursive method

This method solves the problem with a few less lines of code, with the use of recursion to move through the list nodes.

func addTwoNumbers(_ l1: ListNode?, _ l2: ListNode?) -> ListNode? {
    // Start by calling the helper with the initial carry (zero)
    addTwoNumbers(l1, l2, 0)
}

private func addTwoNumbers(_ l1: ListNode?, _ l2: ListNode?, _ carry: Int) -> ListNode? {
    // Base case: return nil if we reach the end of both lists and carry is zero
    guard l1 != nil || l2 != nil || carry > 0 else { return nil }

    // Sum of the current values and carry
    let sum = (l1?.val ?? 0) + (l2?.val ?? 0) + carry
    // New node with the unit digit
    let resultNode = ListNode(sum % 10)
    // Recursively add the next node with the new carry
    resultNode.next = addTwoNumbers(l1?.next, l2?.next, sum / 10)

    // Return the resulting node
    return resultNode
}

Source code

This solution looks pretty good, with just a few lines of code. But can we have an even better solution? ๐Ÿค“

Recursive method (improved)

We can improve that previous solution if we could stop the recursion earlier, once we reach the end of either one of the lists.
Let me show you how to do it. ๐Ÿ‘จโ€๐Ÿซ

func addTwoNumbers(_ l1: ListNode?, _ l2: ListNode?) -> ListNode? {
    // Base cases: return the other list if one is nil (empty)
    guard let l1 else { return l2 }
    guard let l2 else { return l1 }
    
    // Sum of current nodes
    let sum = l1.val + l2.val
    // New node with the unit digit
    let resultNode = ListNode(sum % 10)
    // Recursively add the next node
    let nextNode = addTwoNumbers(l1.next, l2.next)
    if sum > 9 {
        // If the sum is greater than 9, carry 1 to the new node
        resultNode.next = addTwoNumbers(nextNode, ListNode(1))
    } else {
        // Otherwise, there is no carry
        resultNode.next = nextNode
    }
    
    // Return the resulting node
    return resultNode
}

Source code

Complexity Analysis

Time Complexity

Both iterative and recursive methods have the same time complexity: O(max(m, n)), where m and n are the number of nodes in each list. This is because they both go through all of the nodes in both lists, one by one, at the same time. The “improved recursive” version works similar, but will stop earlier in case one list is shorter, so the time complexity is O(min(m, n)). However, if both lists have the same size, all solutions will end with the same O(n) (since m == n in this case).

Space Complexity

All the solutions use the same O(1) constant space. This is because we are only manipulating pointers and not creating any extra space. The total space for each solutions is O(max(m, n)). In some cases, also O(max(m, n) + 1), if there will be a carry at the end.

Conclusion

All solutions described here, whether they are iterative or recursive, are good enough providing a comprehensive understanding of different approaches to solving this problem.

The “improved recursive” version might have a slight advantage when it comes to time complexity, since it could finish earlier.
But in worst case scenarios (when both lists are the same size, for instance), they all will have the same time and space complexities.
Also, while the recursive method handles the problem elegantly by breaking it down into smaller subproblems, it has potential drawbacks, particularly concerning the recursion depth and stack overflow.

So keep all these things in mind when experimenting with these different options. ๐Ÿง‘โ€๐Ÿ”ฌ

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