LeetCode 23. Merge k Sorted Lists

Description

https://leetcode.com/problems/merge-k-sorted-lists/

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 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 <= 10^4
• 0 <= lists[i].length <= 500
• -10^4 <= lists[i][j] <= 10^4
• lists[i] is sorted in ascending order.
• The sum of lists[i].length won’t exceed 10^4.

Explanation

merge sort

Python Solution

# 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[ListNode]) -> ListNode:
        if not lists:
            return None
        
        return self.merge_lists(lists, 0, len(lists) - 1)
    
    def merge_lists(self, lists, start, end):
        if start == end:
            return lists[start]
        
        mid = (start + end) // 2
        left = self.merge_lists(lists, start, mid)
        right = self.merge_lists(lists, mid + 1, end)
        
        return self.merge_two_lists(left, right)
    
    def merge_two_lists(self, head1, head2):
        current = dummy = ListNode(0)
        
        while head1 and head2:
            if head1.val < head2.val:
                current.next = head1
                head1 = head1.next
            else:
                current.next = head2
                head2 = head2.next                
            current = current.next
            
        if head1:
            current.next = head1
        if head2:
            current.next = head2

        return dummy.next    
        

• Time Complexity: ~Nlog(K)
• Space Complexity: ~1