Merge k Sorted Lists

Problem

You are given an array of k linked-lists lists, each linked-list is sorted in ascending order.

Merge all the linked-lists into one sorted linked-list and return it.

Example 1:

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Input: lists = [[1,4,5],[1,3,4],[2,6]]
Output: [1,1,2,3,4,4,5,6]
Explanation: The linked-lists are:
[
  1->4->5,
  1->3->4,
  2->6
]
merging them into one sorted list:
1->1->2->3->4->4->5->6

Example 2:

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Input: lists = []
Output: []

Example 3:

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Input: lists = [[]]
Output: []

Intuition

The given problem requires merging k sorted linked lists into a single sorted linked list. One way to approach this problem is to repeatedly merge pairs of linked lists until there is only one linked list remaining.

Approach

The mergeKLists function takes a list of linked lists as input.

It repeatedly merges pairs of linked lists until there is only one linked list remaining in the list.

The mergelists function is a helper function that merges two sorted linked lists. It iterates through the nodes of both lists, comparing the values, and building a new sorted list.

The merged lists are stored in the mergelist variable, and the process is repeated until there is only one list left.

The final merged list is returned.

Complexity

Time complexity:

The time complexity is O(n log(k)), where n is the total number of nodes across all lists, and k is the number of lists.

Space complexity:

The space complexity is O(1) as we use only a constant amount of extra space.

Code

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def mergeKLists(self, lists: List[Optional[ListNode]]) -> Optional[ListNode]:
        if not lists or len(lists) == 0:
            return None

        while len(lists) > 1:
            mergelist = []

            for i in range(0 , len(lists) , 2):
                l1 = lists[i]
                l2 = lists[i + 1] if i + 1 < len(lists) else None
                mergelist.append(self.mergelists(l1 , l2))

            lists = mergelist

        return lists[0]


    def mergelists(self, l1, l2):
        current = dummy = ListNode()

        while l1 and l2:
            if l1.val < l2.val:
                current.next = l1
                l1 = l1.next
            else:
                current.next = l2
                l2 = l2.next

            current = current.next

        if l1 or l2:
            current.next = l1 if l1 else l2
        
        return dummy.next