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解法一：按顺序两两合并升序列表。先排序lists[0]和lists[0]得到结果result，再将结果result与list[3]排序，以此类推。

/*** 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) {if(lists.length==0){return null;}// 按顺序两两合并升序列表ListNode result = lists[0];for(int i=1;i<lists.length;i++){result = mergeTwoLists(result, lists[i]);}return result;}public ListNode mergeTwoLists(ListNode list1, ListNode list2) {ListNode new_list=null;if(list1 != null && list2 != null){if(list1.val<=list2.val){new_list=list1;list1=list1.next;new_list.next=null;}else{new_list=list2;list2=list2.next;new_list.next=null;}}else{return list1!=null?list1:list2;}ListNode curr=new_list;while(list1 != null && list2 != null){if(list1.val<=list2.val){ListNode temp = list1.next;curr.next=list1;list1.next=null;list1=temp;}else{ListNode temp = list2.next;curr.next=list2;list2.next=null;list2=temp;}curr=curr.next;}if(list1!=null){curr.next=list1;}if(list2!=null){curr.next=list2;}return new_list;}
}

注意：

• 要先判断lists.length==0