Naive Approach: Brute Force

One straightforward approach is to collect all the values from the linked lists into a single array, sort the array, and then create a new linked list from the sorted array.

Algorithm:

1. Iterate through each linked list in the input lists.
2. Extract all the values from each linked list and store them in a single array.
3. Sort the array in ascending order.
4. Create a new linked list from the sorted array.

Code (Python):

class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

def mergeKLists(lists):
    values = []
    for linked_list in lists:
        current = linked_list
        while current:
            values.append(current.val)
            current = current.next
    
    values.sort()
    
    if not values:
        return None
    
    head = ListNode(values[0])
    current = head
    for i in range(1, len(values)):
        current.next = ListNode(values[i])
        current = current.next
        
    return head

Time Complexity: O(N log N), where N is the total number of nodes in all linked lists. This is due to the sorting step.

Space Complexity: O(N), where N is the total number of nodes. This is because we store all node values in an array.

Optimal Approach: Priority Queue (Min-Heap)

A more efficient approach is to use a priority queue (min-heap) to keep track of the smallest node from all the linked lists. This allows us to merge the lists in sorted order without explicitly sorting all the values at once.

Algorithm:

1. Create a min-heap and insert the head nodes of all non-empty linked lists into the heap. The heap should be ordered by the node values.
2. Create a dummy head for the merged linked list.
3. While the heap is not empty:
   • Extract the node with the smallest value from the heap.
   • Append this node to the merged linked list.
   • If the extracted node has a next node, insert the next node into the heap.
4. Return the next of the dummy head, which is the head of the merged linked list.

Code (Python):

import heapq

class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

def mergeKLists(lists):
    heap = []
    for node in lists:
        if node:
            heapq.heappush(heap, (node.val, node))
    
    dummy = ListNode()
    current = dummy
    
    while heap:
        val, node = heapq.heappop(heap)
        current.next = node
        current = current.next
        
        if node.next:
            heapq.heappush(heap, (node.next.val, node.next))
            
    return dummy.next

Time Complexity: O(N log k), where N is the total number of nodes in all linked lists and k is the number of linked lists. Each insertion and extraction from the heap takes O(log k) time, and we perform these operations for each of the N nodes.

Space Complexity: O(k), where k is the number of linked lists. This is the space used by the priority queue.

Edge Cases:

• Empty input list: If the input lists is empty, return None.
• Empty linked lists: If any of the linked lists are empty, they should be skipped when initializing the priority queue. This is handled in the code by checking if node: before pushing onto the heap.
• All empty linked lists: If all linked lists are empty, the priority queue will be empty from the start, and the algorithm will correctly return an empty linked list.