解題思路：
1 之前的某道題有將兩個有序連結串列排序的解法，思路很簡單，可以拿來直接用。
2 利用分治的思想，分根據下標來分；兩兩合併，再兩兩合併…以此類推。

程式碼：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public static ListNode mergeKLists(ListNode[] lists){
        return partion(lists,0,lists.length-1);
    }

    public static ListNode partion(ListNode[] lists,int s,int e){
        if(s==e)  return lists[s];
        if(s<e){
            int q=(s+e)/2;
            ListNode l1=partion(lists,s,q);
            ListNode l2=partion(lists,q+1,e);
            return mergeTwoLists(l1,l2);
        }else
            return null;
    }
 public static ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        ListNode p = l1;
        ListNode q = l2;
        ListNode result;
        ListNode r;
        
        if (l1 == null)
            return l2;
        if (l2 == null)
            return l1;
        
        if (p.val <= q.val){
            result = new ListNode(p.val);
            r = result;
            p = p.next;
        }
        else{
            result = new ListNode(q.val);
            r = result;
            q = q.next;
        }


        while (p != null && q != null){
            while (p != null && q != null && p.val <= q.val){
                r.next = p;
                r = r.next;
                p = p.next;
            }
  
            while (p != null && q != null && q.val <= p.val){
                r.next = q;
                r = r.next;
                q = q.next;
            }
        }
        if (q == null)
            r.next = p;
        else r.next = q;
        
        return result;
    }
}