题解 | #合并k个已排序的链表#

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
        // 优先队列 小根堆 (a,b) -> (a.val - b.val)
        // 大根堆 (b,a) -> (b.val - a.val)
        PriorityQueue<ListNode> queue = new PriorityQueue<>((a,b) -> (a.val - b.val));
        for(ListNode list: lists){
            if(list != null) queue.offer(list);
        }
        ListNode dummyNode = new ListNode(0);
        ListNode cur = dummyNode;
        while(!queue.isEmpty()){
            ListNode temp = queue.poll();
            cur.next = temp;
            cur = cur.next;
            if(temp.next != null){
                queue.offer(temp.next);
            }
        }

        return dummyNode.next;
    }
}