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

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Link for the Problem – Merge 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