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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.

/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode() {} * ListNode(int val) { this.val = val; } * ListNode(int val, ListNode next) { this.val = val; this.next = next; } * } */ class Solution { public ListNode mergeKLists(ListNode[] lists) { int len = lists.length; if(len == 0){ return null; } PriorityQueue<ListNode> pq = new PriorityQueue<ListNode>((a,b)-> a.val - b.val); for(int i=0;i<len;i++){ if(lists[i] == null){ continue; } pq.add(lists[i]); } ListNode dummy = new ListNode(0); ListNode curr = dummy; while(!pq.isEmpty()){ ListNode node = pq.remove(); curr.next = node; curr = curr.next; if(node.next != null){ pq.add(node.next); } } return dummy.next; } }

We can use Divide and Conquer as well and Priority Queue-based approach to solving this problem. A Detailed explanation is provided in the video.

Time Complexity: O(nlogk)

Space Complexity: O(k)