You are given an array, lists, containing k singly linked lists. Each of these linked lists is individually sorted in ascending order.

Your task is to merge all k linked lists into a single sorted linked list in ascending order and return the merged list.

Constraints:

• k $==$ lists.length

• $0 \leq$ k $\leq 10^3$

• $0 \leq$ lists[i].length $\leq 500$

• $-10^3 \leq$ lists[i][j] $\leq 10^3$

• Each lists[i] is sorted in ascending order.

• The sum of all lists[i].length will not exceed $10^3$.

This approach uses a divide-and-conquer method to merge multiple sorted lists into a single sorted list. It initially pairs up the lists and merges each pair. This merging of pairs produces $(\frac{k}{2})$ lists which, when merged, produce $(\frac{k}{4})$ lists, and so on. This is repeated until we’re left with only one fully merged list. Merging a pair of lists uses two pointers, each pointing to a node in one of the two input lists. At each step, the smaller of the two nodes is appended to a new list that holds the merged result. The pointer in the selected list is then moved forward. A dummy head node is introduced as a fixed starting point to make constructing this merged result list easier. New nodes are consistently attached to this dummy-based list, and once the merging is complete, we return the node following the dummy as the head of the final merged list.

Using the intuition above, we implement the algorithm as follows:

1. Check if the input lists is non-empty. If it is:

   1. Initialize a variable step to 1 to control the merging interval.

   2. While step is less than the number of lists, start the pairwise merging process:

      1. Iterate through the list indices from 0 to len(lists) - step, with increments of step * 2:

         1. For each index i, merge the two lists at positions i and i + step using the merge_2_lists helper function.

         2. Replace the list at index i with the merged result.

      2. After finishing one pass of pairwise merges, double the step value to prepare for the next level of merging.

   3. After the outer loop terminates, return the final list stored at lists[0].

The merge_2_lists helper function takes the heads of two linked lists, head1 and head2, and merges the two lists as follows:

1. Create a dummy node to serve as the starting point of the merged list.

2. Initialize a pointer, prev, starting at the dummy node to track where to attach the next node.

3. While both input lists are non-empty:

   1. Compare the current nodes, head1 and head2, and append the smaller node to the merged list using prev.next.

   2. Move the pointer in the selected list forward.

   3. Advance the prev pointer.

4. Once one of the lists is exhausted, attach the remaining nodes from the other list to the merged list.

5. Return the merged list starting at dummy.next, which skips the placeholder dummy node.

Let’s look at the following illustration to get a better understanding of the solution: