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Asked: 2024-09-24T11:42:19+05:30 In:

You are tasked with manipulating linked lists in C. Given two linked lists, each representing a positive integer with its digits stored in reverse order, your goal is to compute the sum of these two integers and return the result as a new linked list, also in reverse order. Each node in the linked list contains a single digit. Implement a function that takes the heads of the two linked lists as input and returns the head of the resultant linked list representing the summation. Ensure that your solution handles edge cases, such as lists of different lengths or empty lists. Provide a well-structured implementation and consider optimal time and space complexity for your solution.

I’m working on a problem involving linked lists in C, and I could really use some help or insight from anyone experienced with this. So, here’s the challenge: imagine you have two linked lists, where each list represents a positive integer, but here’s the twist—they’re stored in reverse order! For example, the number 123 would be represented as 3 -> 2 -> 1.

The task is to create a function that takes in the heads of these two linked lists, adds the numbers they represent, and returns a new linked list as the sum, again in reverse order. So, if you had two linked lists representing the numbers 342 (2 -> 4 -> 3) and 465 (5 -> 6 -> 4), the resulting linked list should represent 807 (7 -> 0 -> 8).

Now, I know this sounds straightforward, but I’ve been hitting some bumps along the way. I have to make sure my function efficiently handles cases like different lengths of lists or even empty ones—what if one of the lists is null? I want to produce a clean, efficient solution because I’m mindful of optimal time and space complexities.

Here are a few scenarios I’ve considered:

1. **Different Lengths**: What if one list has more digits than the other? How do we ensure that all digits are summed correctly?
2. **Carry Handling**: When the sum of two digits exceeds 9, we need to handle the carry to the next digit properly.
3. **Empty Lists**: If either of the input lists is empty, the output should just be the sum of the non-empty list, or zero if both are empty.

I’m trying to figure out the best way to traverse both lists simultaneously, add the corresponding digits, and keep track of carries. If anyone has a neat solution or tips on handling these edge cases, I’d love to hear about it! Plus, sharing an actual implementation or pseudocode would be super helpful!

Thanks in advance for any guidance you can provide!

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    2 Answers


    1. To tackle the problem of adding two linked lists that represent numbers in reverse order, you can create a function that traverses both lists simultaneously, handling different lengths and carries efficiently. First, initialize a new linked list to store the result and a variable to manage the carry from one digit to the next. As you iterate through both lists, sum the values of the nodes along with any carry. If one list is shorter, you can continue adding the remaining digits of the longer list while handling the carry. Additionally, make sure to account for cases where both lists are empty or one of them is null, returning an appropriate result if necessary.

      Here is a sample implementation in C:

      struct ListNode {
    int val;
    struct ListNode *next;
};

struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2) {
    struct ListNode *dummyHead = malloc(sizeof(struct ListNode));
    dummyHead->next = NULL;
    struct ListNode *current = dummyHead;
    int carry = 0;

    while (l1 != NULL || l2 != NULL || carry > 0) {
        int sum = carry;
        if (l1 != NULL) {
            sum += l1->val;
            l1 = l1->next;
        }
        if (l2 != NULL) {
            sum += l2->val;
            l2 = l2->next;
        }
        carry = sum / 10;
        current->next = malloc(sizeof(struct ListNode));
        current->next->val = sum % 10;
        current = current->next;
    }
    current->next = NULL;
    struct ListNode *result = dummyHead->next;
    free(dummyHead);
    return result;
}



    2. Linked List Addition in C

      Sounds like a fun challenge with linked lists! Here’s a simple approach you could take to solve the problem.

      Key Points to Consider:

      • Different Lengths: You can loop through both lists and just treat the missing nodes as zero.
      • Carry Handling: Keep a variable for carry. When the sum of two digits (plus carry) exceeds 9, update the carry for the next iteration.
      • Empty Lists: If a list is empty, just treat it as zero during the addition.

      Pseudocode Overview:

      function addTwoNumbers(l1, l2):
    carry = 0
    dummy_head = new ListNode(0)
    current = dummy_head

    while l1 != null or l2 != null or carry > 0:
        val1 = (l1 != null) ? l1.val : 0
        val2 = (l2 != null) ? l2.val : 0
        
        sum = val1 + val2 + carry
        carry = sum / 10
        current.next = new ListNode(sum % 10)
        current = current.next
        
        if l1 != null: l1 = l1.next
        if l2 != null: l2 = l2.next

    return dummy_head.next

      Explanation:

      This function works as follows:

      1. Initialize a dummy head for the result list and a variable for the carry.
      2. Traverse both lists while there’s at least one node left to process or there’s a carry to add.
      3. If a list is null, consider its value as zero.
      4. Compute the sum of the two digits and the carry, then create a new node for the digit part of the sum.
      5. Move to the next node in both lists.

      With this method, you’ll efficiently handle different lengths, carries, and empty lists. Hope this helps you out!


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