LC 23 - Merge k Sorted Lists

Question

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:

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 linked list: 1->1->2->3->4->4->5->6

Example 2:

Input: lists = []
Output: []

Example 3:

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

Constraints:

• k == lists.length
• 0 <= k <= 104
• 0 <= lists[i].length <= 500
• -104 <= lists[i][j] <= 104
• lists[i] is sorted in ascending order.
• The sum of lists[i].length will not exceed 104.

Links

Question here and solution here

Solution

concept

This question is an extension to [[21. Merge Two Sorted Lists]], we basically iterate through the list of linked list and merge each pair using the solution from the previous questions to merge two list together

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:
            _merged_list = [] # tmp variable

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

    def merge_lists(self, l1, l2):
	    """
	    solution of 21. Merge Two Sorted Lists
	    """
        dummy = ListNode()
        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
        
        if l1:
            tail.next = l1
        elif l2:
            tail.next = l2
        
        return dummy.next

Complexity

time: $O(n*k)$, k is the total number of lists and n is the total number of nodes across k lists.
space: $O(1)$