Leetcode merge K sorted lists solution report

Https://oj.leetcode.com/problems/merge-k-sorted-lists/
Merge K sorted arrays and analyze the complexity of the entire algorithm.

Solution report: the simplest implementation method is to traverse the list <listnode>, perform a merge operation similar to the last step of merging and sorting for each linked list and the current linked list.
The algorithm complexity is O (kN)

public class Solution {    ListNode mergeTwoLists(ListNode list1, ListNode list2) {        ListNode head    = new ListNode(-1);        ListNode current = head;        while(list1!=null&&list2!=null) {            if(list1.val<list2.val) {                current.next = list1;                list1   = list1.next;            } else {                current.next = list2;                list2   = list2.next;            }            current = current.next;        }        if(list1!=null) {            current.next = list1;        } else {            current.next = list2;        }        return head.next;    }    public ListNode mergeKLists(List<ListNode> lists) {        if(lists==null||lists.size()==0) {            return null;        }        ListNode head = lists.get(0);        for(int i=1;i<lists.size();i++) {            head = mergeTwoLists(head, lists.get(i));        }        return head;    }}

The above method TLE, I checked it online and noticed that by using the Merge Sorting Algorithm, the time complexity of the chain table sorting can be reduced to O (nlgn). The specific calculation formula is:

Therefore, the following code is successfully AC by referring to the merge recursion method. Time Complexity: O (nlogk)

public class Solution {    ListNode merge2Lists(ListNode list1, ListNode list2) {        ListNode head    = new ListNode(-1);        ListNode current = head;        while(list1!=null&&list2!=null) {            if(list1.val<list2.val) {                current.next = list1;                list1   = list1.next;            } else {                current.next = list2;                list2   = list2.next;            }            current = current.next;        }        if(list1!=null) {            current.next = list1;        } else {            current.next = list2;        }        return head.next;    }    public ListNode mergeKLists(List<ListNode> lists) {        if(lists==null||lists.size()==0) {            return null;        }        if(lists.size()==1) {            return lists.get(0);        }        int length = lists.size() ;        int mid = (length - 1)/2 ;        ListNode l1 = mergeKLists(lists.subList(0,mid + 1)) ;        ListNode l2 = mergeKLists(lists.subList(mid + 1,length)) ;        return merge2Lists(l1,l2) ;    }}

Leetcode merge K sorted lists solution report