Merge k Sorted Lists

Question

Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.

Method

Use the thought of merging and sorting to merge the data until it finally becomes one.

    public ListNode mergeKLists(ArrayList
  
    lists) {        int len = lists.size();        if(len == 0){            return null;        }        int n = ( len + 1 ) / 2;        while(len > 1){            for(int i = 0; i < n ; i++){                if(n + i < len){                    if(lists.get(i) == null){                        lists.set(i, lists.get(n + i));                        lists.remove(n + i);                    }else if(lists.get(n + i) == null){                        lists.remove(n + i);                    }else{                        ListNode first = lists.get(i);                        ListNode second = lists.get(n + i);                        ListNode head = null;                        ListNode currentNode = null;                        ListNode node;                        while(first != null && second != null){                            if(first.val < second.val){                                node = first;                                first = first.next;                            }else{                                node = second;                                second = second.next;                            }                            if(head == null){                                head = node;                                currentNode = head;                            }else{                                currentNode.next = node;                                node.next = null;                                currentNode = node;                            }                        }                        if(first != null){                            currentNode.next = first;                        }else{                            currentNode.next = second;                        }                         lists.set(i, head);                        lists.remove(n + i);                                            }                }                len = lists.size();                n = (len + 1) / 2;            }        }        return lists.get(0);    }