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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& lists) { ListNode dummy(0); ListNode* curr = &dummy; auto compare = [](const ListNode* a, const ListNode* b) { return a->val > b->val; // min-heap }; priority_queue , 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; Queuepq = 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