Easy Linked Lists ⭐ Standard Interview

Merge Two Sorted Linked Lists — Dummy Node Technique (Python, C, JS)

Problem Statement

Given heads of two sorted linked lists, merge them into one sorted list and return its head.

Example

Input: list1 = [1->2->4], list2 = [1->3->4]

Output: [1->1->2->3->4->4]

Time Complexity

O(n + m)

Space Complexity

O(1)

The Intuition

Dummy head node to avoid head-initialisation edge cases. Compare front nodes, attach smaller, advance that pointer.

Solution Code

def mergeTwoLists(l1, l2):
    dummy = ListNode(0)
    tail = dummy
    while l1 and l2:
        if l1.val < l2.val:
            tail.next = l1; l1 = l1.next
        else:
            tail.next = l2; l2 = l2.next
        tail = tail.next
    tail.next = l1 or l2
    return dummy.next

struct ListNode* mergeTwoLists(struct ListNode* l1, struct ListNode* l2) {
    struct ListNode dummy; struct ListNode* tail = &dummy; dummy.next = NULL;
    while (l1 && l2) {
        if (l1->val < l2->val) { tail->next = l1; l1 = l1->next; }
        else { tail->next = l2; l2 = l2->next; }
        tail = tail->next;
    }
    tail->next = l1 ? l1 : l2;
    return dummy.next;
}

function mergeTwoLists(l1, l2) {
    const dummy = new ListNode(0);
    let tail = dummy;
    while (l1 && l2) {
        if (l1.val < l2.val) { tail.next = l1; l1 = l1.next; }
        else { tail.next = l2; l2 = l2.next; }
        tail = tail.next;
    }
    tail.next = l1 || l2;
    return dummy.next;
}

Frequently Asked Questions

Q: Why use a dummy node?

It avoids special-case logic for initialising the head of the merged list — tail.next handles every node uniformly.

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