leetcode 23. 合并K个排序链表

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

代码：

    两两合并法：

public ListNode mergeKLists(ListNode[] lists) {
        ListNode result = null;
        for(ListNode listNode : lists){
            result = merged(result,listNode);
        }
        return result;
    }

    public ListNode merged(ListNode one , ListNode two){
        if(one==null)
            return two;
        if(two==null)
            return one;
        ListNode head = new ListNode(0);
        ListNode tmp = head;
        while (one!=null&&two!=null){
            if(one.val<=two.val){
                tmp.next = one;
                ListNode tmpNode = one.next;
                one.next = null;
                one = tmpNode;
                tmp = tmp.next;
            }
            else {
                tmp.next = two;
                ListNode tmpNode = two.next;
                two.next = null;
                two = tmpNode;
                tmp = tmp.next;
            }
        }
        if(one!=null){
            tmp.next = one;
        }
        if(two!=null){
            tmp.next = two;
        }
        return head.next;
    }

归并合并法：

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 start ,int end){
        if(start>=end)
            return lists[start];
        int middle = (start+end)/2;
        ListNode one = merge(lists,start,middle);
        ListNode two = merge(lists,middle+1,end);
        return merged(one,two);
    }

    public ListNode merged(ListNode one , ListNode two){
        if(one==null)
            return two;
        if(two==null)
            return one;
        ListNode head = new ListNode(0);
        ListNode tmp = head;
        while (one!=null&&two!=null){
            if(one.val<=two.val){
                tmp.next = one;
                ListNode tmpNode = one.next;
                one.next = null;
                one = tmpNode;
                tmp = tmp.next;
            }
            else {
                tmp.next = two;
                ListNode tmpNode = two.next;
                two.next = null;
                two = tmpNode;
                tmp = tmp.next;
            }
        }
        if(one!=null){
            tmp.next = one;
        }
        if(two!=null){
            tmp.next = two;
        }
        return head.next;
    }