Merge k Sorted Lists LeetCode Programming Solutions | LeetCode Problem Solutions in C++, Java, & Python [💯Correct]

Merge k Sorted Lists LeetCode Problem | LeetCode Problems For Beginners | LeetCode Problems & Solutions | Improve Problem Solving Skills | LeetCode Problems Java | LeetCode Problems C++

Hello Programmers/Coders, Today we are going to share solutions to the Programming problems of LeetCode Solutions in C++, Java, & Python. At Each Problem with Successful submission with all Test Cases Passed, you will get a score or marks and LeetCode Coins. And after solving maximum problems, you will be getting stars. This will highlight your profile to the recruiters.

In this post, you will find the solution for the Merge k Sorted Lists in C++, Java & Python-LeetCode problem. We are providing the correct and tested solutions to coding problems present on LeetCode. If you are not able to solve any problem, then you can take help from our Blog/website.

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About LeetCode

LeetCode is one of the most well-known online judge platforms to help you enhance your skills, expand your knowledge and prepare for technical interviews. 

LeetCode is for software engineers who are looking to practice technical questions and advance their skills. Mastering the questions in each level on LeetCode is a good way to prepare for technical interviews and keep your skills sharp. They also have a repository of solutions with the reasoning behind each step.

LeetCode has over 1,900 questions for you to practice, covering many different programming concepts. Every coding problem has a classification of either EasyMedium, or Hard.

LeetCode problems focus on algorithms and data structures. Here is some topic you can find problems on LeetCode:

  • Mathematics/Basic Logical Based Questions
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  • Stack & Queue
  • Trees & Graphs
  • Greedy Algorithms
  • Breadth-First Search
  • Depth-First Search
  • Sorting & Searching
  • BST (Binary Search Tree)
  • Database
  • Linked List
  • Recursion, etc.

Leetcode has a huge number of test cases and questions from interviews too like Google, Amazon, Microsoft, Facebook, Adobe, Oracle, Linkedin, Goldman Sachs, etc. LeetCode helps you in getting a job in Top MNCs. To crack FAANG Companies, LeetCode problems can help you in building your logic.

Link for the ProblemMerge k Sorted Lists– LeetCode Problem

Merge k Sorted Lists– LeetCode 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:

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

Example 2:

Input: lists = []
Output: []

Example 3:

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


  • 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.
Merge k Sorted Lists– LeetCode Solutions
class Solution {
  ListNode* mergeKLists(vector<ListNode*>& lists) {
    ListNode dummy(0);
    ListNode* curr = &dummy;
    auto compare = [](const ListNode* a, const ListNode* b) {
      return a->val > b->val;  // min-heap
    priority_queue<ListNode*, vector<ListNode*>, decltype(compare)> pq(compare);

    for (ListNode* list : lists)
      if (list)

    while (!pq.empty()) {
      ListNode* minNode =;
      if (minNode->next)
      curr->next = minNode;
      curr = curr->next;

class Solution {
  public ListNode mergeKLists(ListNode[] lists) {
    ListNode dummy = new ListNode(0);
    ListNode curr = dummy;
    Queue<ListNode> pq = new PriorityQueue<>((a, b) -> a.val - b.val); // min-heap

    for (final ListNode list : lists)
      if (list != null)

    while (!pq.isEmpty()) {
      ListNode minNode = pq.poll();
      if ( != null)
        pq.offer(; = minNode;
      curr =;

from queue import PriorityQueue

class Solution:
  def mergeKLists(self, lists: List[ListNode]) -> ListNode:
    dummy = ListNode(0)
    curr = dummy
    pq = PriorityQueue()

    for i, lst in enumerate(lists):
      if lst:
        pq.put((lst.val, i, lst))

    while not pq.empty():
      _, i, minNode = pq.get()
        pq.put((, i, = minNode
      curr =


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