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

对每一个排序链表都设置一个指针。每次讲指针指向的单元中最小值链接，直到所有指针为空。

public class Solution {
   public ListNode mergeKLists(List<ListNode> lists) {
       
        int length = lists.size();
        if(length == 0)return null;
        int num = 0;
        while(num<length)
        {
            if(lists.get(num)!=null)
                break;
            num++;
        }
        if(num == length)return null;
        ListNode[] start = new ListNode[length];
        int minIndex = 0;
        int min = Integer.MAX_VALUE;
        for(int i=0;i<length;i++)
        {
            start[i] = lists.get(i);
            if(start[i]!=null&&start[i].val<min)
            {
                minIndex = i;
                min = start[i].val;
            }
        }
        ListNode head = lists.get(minIndex);
        ListNode node = head;
        start[minIndex] = start[minIndex].next;
        while(!isAllEnd(start))
        {
            findMin(start,node);
            node = node.next;
        }
       
        return head;
       
       
    }
    public void findMin(ListNode[] lists,ListNode node)
    {
        int min = Integer.MAX_VALUE;
        int minIndex = 0;
        for(int i=0;i<lists.length;i++)
        {
            if(lists[i]!=null&&lists[i].val<min)
            {
                min = lists[i].val;
                minIndex = i;
            }
        }
        node.next = lists[minIndex];
        lists[minIndex] = lists[minIndex].next;
    }
   
    public boolean isAllEnd(ListNode[] lists)
    {
        int length = lists.length;
        int i = 0;
        while(i<length)
        {
            if(lists[i] != null)
            {
                return false;
            }
            i++;
        }
        return true;
    }
}