【力扣刷题】23. 合并K个升序链表

输入：lists = [[1,4,5],[1,3,4],[2,6]]
输出：[1,1,2,3,4,4,5,6]
解释：链表数组如下：
[
  1->4->5,
  1->3->4,
  2->6
]
将它们合并到一个有序链表中得到。
1->1->2->3->4->4->5->6

输入：lists = []
输出：[]

输入：lists = [[]]
输出：[]

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode mergeKLists(ListNode[] lists) {
          //新建一个头节点
        ListNode res = new ListNode(0);
        //当前节点指针
        ListNode cur = res;
        List<Integer> list = new ArrayList<>();
        for (ListNode listNode : lists) {
            while (listNode != null) {
                list.add(listNode.val);
                listNode = listNode.next;
            }
        }
        List<Integer> collect = list.stream().sorted().collect(Collectors.toList());
        for (int i = 0; i < collect.size(); i++) {
            ListNode listNode = new ListNode(collect.get(i));
            cur.next = listNode;
            cur = cur.next;
        }
        return res.next;
    }
}