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

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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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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

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:
[
  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.
Merge k Sorted Lists– LeetCode Solutions
class Solution {
 public:
  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)
        pq.push(list);

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

    return dummy.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)
        pq.offer(list);

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

    return dummy.next;
  }
}
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()
      if minNode.next:
        pq.put((minNode.next.val, i, minNode.next))
      curr.next = minNode
      curr = curr.next

    return dummy.next

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