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23. Merge k Sorted Lists 👍

  • Time: $O(n\log k)$
  • Space: $O(k)$
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class Solution {
 public:
  ListNode* mergeKLists(vector<ListNode*>& lists) {
    ListNode dummy(0);
    ListNode* curr = &dummy;
    auto compare = [](ListNode* a, ListNode* b) { return a->val > b->val; };
    priority_queue<ListNode*, vector<ListNode*>, decltype(compare)> minHeap(
        compare);

    for (ListNode* list : lists)
      if (list != nullptr)
        minHeap.push(list);

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

    return dummy.next;
  }
};

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class Solution {
  public ListNode mergeKLists(ListNode[] lists) {
    ListNode dummy = new ListNode(0);
    ListNode curr = dummy;
    Queue<ListNode> minHeap = new PriorityQueue<>((a, b) -> Integer.compare(a.val, b.val));

    for (final ListNode list : lists)
      if (list != null)
        minHeap.offer(list);

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

    return dummy.next;
  }
}

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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