The is problem could be elegantly solved by using merge.The inefficient way is to scan the lists from the first list to the end list. However this way is too inefficient, each element in the list needed to be compared i - 1 times. (lists[i])An elegant way is to use the idea from merge sort. We could merge the lists in binary way.To achieve this goal, we need to define the recursion. (It includes some skills)1. we use two indexes: low and high, partition into two parts, then sort and each partition repsectively.    a. iff low < high, we partition the lists into <low .... mid> ,  <mid + 1 ....high>.     b. iff low == high, there are only one list (lists[low]) left in the lists, we could directly return it. (base case)2. merge the two partitions, and return the result as one linked list.

public class Solution {    public ListNode mergeKLists(List<ListNode> lists) {        if (lists == null || lists.size() == 0)            return null;        return helper(lists, 0, lists.size()-1);    }        private ListNode helper(List<ListNode> lists, int start, int end) {        if (start == end)            return lists.get((start + end)/2);        int mid = (start + end)/2;        ListNode l1 = helper(lists, start, mid);        ListNode l2 = helper(lists, mid+1, end);        ListNode new_list = merge(l1, l2);        return new_list;    }        private ListNode merge(ListNode head1, ListNode head2) {        if (head1 == null)            return head2;        if (head2 == null)            return head1;        ListNode l1 = head1;        ListNode l2 = head2;        ListNode dummy = new ListNode(0);        ListNode pre = dummy;        while (l1 != null && l2 != null) {            if (l1.val < l2.val) {                ListNode next1 = l1.next;                l1.next = pre.next;                pre.next = l1;                pre = l1;                l1 = next1;            } else{                ListNode next2 = l2.next;                l2.next = pre.next;                pre.next = l2;                pre = l2;                l2 = next2;            }        }        if (l1 == null)            pre.next = l2;        else             pre.next = l1;        return dummy.next;    }}