题解 | 2025/7/24（链表）（Java）

import java.util.*;

/*
 * public class ListNode {
 *   int val;
 *   ListNode next = null;
 *   public ListNode(int val) {
 *     this.val = val;
 *   }
 * }
 */

public class Solution {
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param lists ListNode类ArrayList 
     * @return ListNode类
     */
    public ListNode mergeKLists (ArrayList<ListNode> lists) {
        // write code here
        ListNode dummy = new ListNode(0);
        PriorityQueue<ListNode> pq = new PriorityQueue<>(Comparator.comparingInt(node -> node.val));
        for (ListNode tmp : lists) {
            if (tmp == null) continue;  // 去除一开始为空
            pq.offer(tmp);
        }
        ListNode cur = dummy;
        while (!pq.isEmpty()) {
            ListNode tmp = pq.poll();
            cur.next = tmp;
            cur = tmp;
            if (tmp.next != null) {
                pq.offer(tmp.next);
            }
        }
        return dummy.next;
    }
}