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.

Problem Statement

The problem is to add two non-empty linked lists representing two non-negative integers. The digits of each integer are stored in reverse order in the linked lists, where each node of the linked lists contains a single digit. The task is to calculate the sum of the two integers and return the result as a linked list with the digits of the sum also stored in reverse order. It can be assumed that the two numbers do not contain any leading zero, except the number 0 itself.

Examples

```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]```

Clarifying Questions

1. Are the linked lists of equal length or can they be of different lengths?
2. Are the digits stored in the linked lists in reverse order (least significant digit first)?
3. Are there any restrictions on the length of the linked lists?
4. Can we assume that the input linked lists are non-empty?
5. Are the digits in each node guaranteed to be single-digit non-negative integers (0-9)?
6. Can we assume that the output linked list should also be in reverse order (least significant digit first)?
7. Are there any other special cases or constraints that I should be aware of?

Solutions

There are different approaches to solving this problem. Here are a few common ones:

1. Iterate through both linked lists and add the corresponding digits, keeping track of any carryover, and building a new linked list with the resulting sum.

2. Convert the linked lists to integers, add them, and then convert the resulting sum back into a linked list.

3. Use recursion to traverse the linked lists and perform the addition, and create a new linked list with the resulting sum.

4. Reverse the linked lists, perform the addition, and then reverse the resulting linked list back to the original order.

Each of these approaches has its advantages and disadvantages, and the best approach depends on the specific requirements and constraints of the problem, as well as the preferences and experience of the developer.

Approach 1: Iterating through linked lists

Iterate through both linked lists and add the corresponding digits, keeping track of any carryover, and building a new linked list with the resulting sum.

Algorithm:

1. Initialize a new empty linked list.
2. Traverse the two input linked lists in parallel, starting from the head nodes, until both have reached the end of the list.
3. At each step, add the corresponding digits of the two linked lists, along with any carryover from the previous addition.
4. If the resulting sum is greater than 9, set the carryover to 1 and add the least significant digit of the sum to the new linked list.
5. If the resulting sum is less than or equal to 9, set the carryover to 0 and add the sum to the new linked list.
6. If one of the linked lists has more digits than the other, continue to add the digits of the longer linked list to the carryover until there are no more digits left.
7. If there is a carryover remaining at the end of the traversal, add it as an additional digit to the new linked list.
8. Return the resulting linked list.

Time complexity: O(max(m, n)), where m and n are the lengths of the input linked lists

Space complexity: O(max(m, n)), the length of the result linked list

Approach 2: Converting linked lists to integers

Convert the linked lists to integers, add them, and then convert the resulting sum back into a linked list.

Algorithm:

1. Traverse each input linked list in reverse order, starting from the tail node, and convert each linked list into an integer using the reversed digit order.
2. Add the two resulting integers.
3. Traverse the resulting integer digits in reverse order and create a new linked list with each digit as a node in the linked list.
4. Return the resulting linked list.

Time complexity: O(max(m, n)), where m and n are the lengths of the input linked lists

Space complexity: O(max(m, n)), the maximum recursion depth

Approach 3: Recursion

Use recursion to traverse the linked lists and perform the addition, and create a new linked list with the resulting sum.

Algorithm:

1. Initialize a new empty linked list.
3. Define a recursive function that takes in two nodes as input, representing the current digits being added, and a carryover value.
4. In the recursive function, add the current digit values of the two nodes, along with the carryover value.
5. If the resulting sum is greater than 9, set the carryover to 1 and create a new node in the resulting linked list with the least significant digit of the sum.
6. If the resulting sum is less than or equal to 9, set the carryover to 0 and create a new node in the resulting linked list with the sum as the value.
7. If either of the input nodes is null, set the value of that node to 0 and continue with the addition.
8. Recursively call the function with the next nodes in each linked list and the new carryover value.
9. Return the resulting linked list.

Time complexity: O(max(m, n)), where m and n are the lengths of the input linked lists

Space complexity: O(max(m, n)), the length of the result linked list

Approach 4: Reversing the linked list

Reverse the linked lists, perform the addition, and then reverse the resulting linked list back to the original order.

Algorithm:

1. Reverse both input linked lists by reversing the order of nodes in each list.
2. Follow the algorithm for approach 1, iterating through both linked lists in parallel, adding corresponding digits, and keeping track of carryover to build a new linked list with the resulting sum.
3. Reverse the resulting linked list to restore the original order of digits.
4. Return the resulting linked list.

Time complexity: O(max(m, n)), where m and n are the lengths of the input linked lists

Space complexity: O(max(m, n)), the length of the result linked list

All four approaches have the same time and space complexity of O(max(m, n)), where m and n are the lengths of the input linked lists. Therefore, all four approaches have the same asymptotic time and space complexity, and none of them is more optimal than the others in terms of Big O notation.

However, in practice, there may be some differences in performance due to the specific implementation details and the underlying hardware. For example, the recursive approach (Approach 3) may have slightly slower performance due to the overhead of the recursive function calls. The other two approaches (Approach 1 and Approach 2) are likely to have similar performance characteristics, but may differ in their implementation details.

Ultimately, the choice of which approach to use will depend on factors such as the specific requirements of the use case, the programming language and libraries being used, and any performance benchmarks or constraints that need to be met.

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