MergeKSortedLists

Problem

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 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 <= 10:sup:`4`

• 0 <= lists[i].length <= 500

• -10:sup:`4`<= lists[i][j] <= 10:sup:`4`

• lists[i] is sorted in ascending order.

• The sum of lists[i].length will not exceed 10:sup:`4`.

Solution

Same as merge 2 sorted linked lists: pick the smallest head and add it to the end of the new list.

Pattern

Linked List, Divide and Conquer, Heap (Priority Queue), Merge Sort

Code

from __future__ import annotations

import heapq
from typing import List


class ListNode:
    """Node in a linked list.
    """

    def __init__(self, val: int, next: ListNode | None = None):
        self.val = val
        self.next = next

    def __lt__(self, node: ListNode) -> bool:
        return self.val < node.val

    def __eq__(self, node: object) -> bool:
        return self.val == node.val

    def __str__(self) -> str:
        return str(self.val) + (' -> ' + str(self.next) if self.next is not None else '')

    @classmethod
    def from_list(cls, list: List[int]) -> ListNode | None:
        head: ListNode | None = None
        prev_node: ListNode | None = None
        for el in list:
            node = ListNode(el)
            if head is None:
                head = node
            elif prev_node is not None:
                prev_node.next = node
                prev_node = node
            prev_node = node
        return head


def merge_k_lists(heads: List[ListNode | None]) -> ListNode | None:
    """Merge k sorted linked lists.
    """
    not_none_heads: List[ListNode] = [head for head in heads if head is not None]
    heapq.heapify(not_none_heads)
    new_head: ListNode | None = None
    prev_node: ListNode | None = None
    while len(not_none_heads) > 0:
        node = heapq.heappop(not_none_heads)
        if new_head is None:
            new_head = node
        elif prev_node is not None:
            prev_node.next = node
        prev_node = node
        if node.next is not None:
            heapq.heappush(not_none_heads, node.next)
    return new_head

Test

>>> from MergeKSortedLists import ListNode, merge_k_lists
>>> head1 = ListNode.from_list([1, 5, 6])
>>> head2 = ListNode.from_list([2, 3, 7])
>>> head3 = ListNode.from_list([2, 4, 8])
>>> head = merge_k_lists([head1, head2, head3])
>>> print(head)
1 -> 2 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8

class MergeKSortedLists.ListNode(val: int, next: ListNode | None = None)

Bases: object

Node in a linked list.

classmethod from_list(list: List[int]) → ListNode | None

MergeKSortedLists.merge_k_lists(heads: List[ListNode | None]) → ListNode | None

Merge k sorted linked lists.