offer2-78

给定一个链表数组，每个链表都已经按升序排列。

请将所有链表合并到一个升序链表中，返回合并后的链表。

示例 1：

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

示例 2：

输入：lists = []
输出：[]

示例 3：

输入：lists = [[]]
输出：[]

提示：

• k == lists.length
• 0 <= k <= 10^4
• 0 <= lists[i].length <= 500
• -10^4 <= lists[i][j] <= 10^4
• lists[i] 按 升序 排列
• lists[i].length 的总和不超过 10^4

//leetcode submit region begin(Prohibit modification and deletion)
/**
 * 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) {
        if (lists == null || lists.length == 0) {
            return null;
        }

        PriorityQueue<ListNode> pq = new PriorityQueue<>(new Comparator<ListNode>(){
            @Override
            public int compare(ListNode node1, ListNode node2) {
                return node1.val - node2.val;
            }
        });

        for (ListNode n : lists) {
            if (n != null) {
                pq.add(n);
            }
        }

        ListNode head = null;
        ListNode pre = null;

        while (!pq.isEmpty()) {
            ListNode node = pq.poll();
            if (head == null) {
                head = node;
                pre = node;
                if (node.next != null) {
                    pq.add(node.next);
                }
                continue;
            }

            pre.next = node;
            if (node.next != null) {
                pq.add(node.next);
            }
            pre = node;
        }

        if (pre != null) {
            pre.next = null;
        }
        return head;
    }
}
//leetcode submit region end(Prohibit modification and deletion)