🚀Day 3. Merge Two Sorted Linked Lists 🧠

The problem can be found at the following link: Problem Link

💡 Problem Description:

Given the heads of two sorted linked lists, the task is to merge the two lists into a single sorted linked list and return the head of the merged list.

🔍 Example Walkthrough:

Example 1:

Input: head1 = 5 -> 10 -> 15 -> 40 head2 = 2 -> 3 -> 20

Output: 2 -> 3 -> 5 -> 10 -> 15 -> 20 -> 40 Explaination:

Example 2:

Input: head1 = 1 -> 1 head2 = 2 -> 4

Output: 1 -> 1 -> 2 -> 4

Constraints:

• 1 <= no. of nodes<= $10^3$

• $0 <= node->data <= 10^5$

🎯 My Approach:

Iterative Approach:

1. Create a dummy node to simplify the merging process and maintain a pointer tail starting at the dummy node.

2. Traverse both linked lists using two pointers, head1 and head2.

   • Compare the data of the current nodes from both lists.

   • Attach the node with the smaller value to tail and move the pointer of the respective list.

3. Once one of the lists is completely traversed, attach the remaining nodes of the other list to tail.

4. Return the merged list starting from dummy.next.

🕒 Time and Auxiliary Space Complexity

• Expected Time Complexity: O(n + m), where n and m are the lengths of the two linked lists. Each node is visited exactly once.

• Expected Auxiliary Space Complexity: O(1), as no additional space is used except for a few variables.

📝 Solution Code

Code (C)

struct Node* sortedMerge(struct Node* head1, struct Node* head2) {
    struct Node dummy = {0, NULL}, *tail = &dummy;
    while (head1 && head2) {
        if (head1->data <= head2->data) {
            tail->next = head1;
            head1 = head1->next;
        } else {
            tail->next = head2;
            head2 = head2->next;
        }
        tail = tail->next;
    }
    tail->next = head1 ? head1 : head2;
    return dummy.next;
}

Code (C++)

Iterative Approach (Tail Insertion Merge)

class Solution {
public:
    Node* sortedMerge(Node* head1, Node* head2) {
        Node dummy(0), *tail = &dummy;
        while (head1 && head2) {
            tail->next = (head1->data <= head2->data) ? head1 : head2;
            tail = tail->next;
            if (head1->data <= head2->data) head1 = head1->next;
            else head2 = head2->next;
        }
        tail->next = head1 ? head1 : head2;
        return dummy.next;
    }
};

👨‍💻 Alternative Approaches

Recursive Merge

• This approach uses recursion to decide which node should be part of the merged list.

1. Compare the data of the nodes at head1 and head2.

2. Recursively determine the next pointer of the smaller node by calling the function with the rest of the nodes.

3. Continue until one of the lists is empty, at which point return the other list.

class Solution {
public:
    Node* sortedMerge(Node* head1, Node* head2) {
        if (!head1) return head2;
        if (!head2) return head1;
        if (head1->data <= head2->data) {
            head1->next = sortedMerge(head1->next, head2);
            return head1;
        }
        head2->next = sortedMerge(head1, head2->next);
        return head2;
    }
};

Iterative Merge with Two Pointers

This approach is the same as described above but focuses on using a pointer swapping technique.

class Solution {
public:
    Node* sortedMerge(Node* head1, Node* head2) {
        Node dummy(0), *tail = &dummy;
        for (; head1 && head2; tail = tail->next)
            tail->next = (head1->data <= head2->data) ? std::exchange(head1, head1->next) : std::exchange(head2, head2->next);
        tail->next = head1 ? head1 : head2;
        return dummy.next;
    }
};

Code (Java)

class Solution {
    Node sortedMerge(Node head1, Node head2) {
        Node dummy = new Node(0), tail = dummy;
        while (head1 != null && head2 != null) {
            tail.next = (head1.data <= head2.data) ? head1 : head2;
            if (head1.data <= head2.data) head1 = head1.next;
            else head2 = head2.next;
            tail = tail.next;
        }
        tail.next = (head1 != null) ? head1 : head2;
        return dummy.next;
    }
}

Code (Python)

class Solution:
    def sortedMerge(self, head1, head2):
        dummy = Node(0)
        tail = dummy
        while head1 and head2:
            if head1.data <= head2.data:
                tail.next = head1
                head1 = head1.next
            else:
                tail.next = head2
                head2 = head2.next
            tail = tail.next
        tail.next = head1 or head2
        return dummy.next

🎯 Contribution and Support:

For discussions, questions, or doubts related to this solution, feel free to connect on LinkedIn: Any Questions. Let’s make this learning journey more collaborative!

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