题目：

合并 k 个排序链表，返回合并后的排序链表。请分析和描述算法的复杂度。

示例:

输入:
[
  1->4->5,
  1->3->4,
  2->6
]
输出: 1->1->2->3->4->4->5->6

我的答案：

解法一：暴力法

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode mergeKLists(ListNode[] lists) {
        //ListNode node = new ListNode(0);
        if(lists==null||lists.length==0){
            return null;
        }
        //node.next = lists.get(0);
        ListNode temp = lists[0];
        for(int i=1;i<lists.length;i++){
            temp = mergeTwoLists(temp,lists[i]);
        }
        return temp;
    }
    public ListNode mergeTwoLists(ListNode l1,ListNode l2){
        ListNode node = new ListNode(0);
        ListNode temp = node;
        while(l1!=null&&l2!=null){
            if(l1.val<=l2.val){
                temp.next = l1;
                l1 = l1.next;
            }else{
                temp.next = l2;
                l2 = l2.next;
            }
            temp = temp.next;
        }
        if(l1!=null){
            temp.next = l1;
        }
        if(l2!=null){
            temp.next = l2;
        }
        return node.next;
    }
}

解法二：分治法

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode mergeKLists(ListNode[] lists) {
        if(lists==null||lists.length==0){
            return null;
        }
        return merge(lists,0,lists.length-1);
    }
    public ListNode merge(ListNode[] lists,int left,int right){
        if(left == right){
            return lists[left];
        }
        int mid = (left+right)/2;
        ListNode l1 = merge(lists,left,mid);
        ListNode l2 = merge(lists,mid+1,right);
        return mergeTwoLists(l1,l2);
    }
    public ListNode mergeTwoLists(ListNode l1,ListNode l2){
        ListNode node = new ListNode(0);
        ListNode temp = node;
        while(l1!=null&&l2!=null){
            if(l1.val<=l2.val){
                temp.next = l1;
                l1 = l1.next;
            }else{
                temp.next = l2;
                l2 = l2.next;
            }
            temp = temp.next;
        }
        if(l1!=null){
            temp.next = l1;
        }
        if(l2!=null){
            temp.next = l2;
        }
        return node.next;
    }
}