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

解析：合并k个已经有序的单链表，使其最终成为一个有序的单链表。原理就是归并排序，递归运算。基本算法recusion 与 merge

编码：

public ListNode mergeKLists(ListNode[] lists) {        if(lists == null || lists.length == 0)           return null;        if(lists.length == 1)           return lists[0];        return recursion(lists,0,lists.length - 1);    }    //recursion    public ListNode recursion(ListNode[] lists,int start,int end){        if(start == end)//只有一个链表           return lists[start];        if(start < end){            int mid = start + (end - start) / 2; //注意：这里防止整数越界的处理,start+(end-start)/2            ListNode l1 = recursion(lists,start,mid);            ListNode l2 = recursion(lists,mid + 1,end);            return merge(l1,l2);        } else            return null;            }    //merge    public ListNode merge(ListNode l1,ListNode l2){        ListNode head = new ListNode(0); //创建一个头结点，最后还要删掉        ListNode p = head;        while(l1 != null && l2 != null){            if(l1.val <= l2.val){                p.next = l1;                l1 = l1.next;            } else{                p.next = l2;                l2 = l2.next;            }            p = p.next;        }                p.next = (l1 != null) ? l1 : l2;        return head.next;// head的下一个节点是第一个数据结点    }