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
        Queue<ListNode> priorityQueue = new PriorityQueue<>((v1,
                v2) -> v1.val - v2.val);//创造小根堆
        for (int i = 0; i < lists.size(); i++) {
            if (lists.get(i) != null) {
                priorityQueue.add(lists.get(i));
            }
        }
        ListNode res = new ListNode(-1);
        ListNode head = res;
        while (!priorityQueue.isEmpty()) {
            ListNode tmp = priorityQueue.poll();
            head.next = tmp;
            head = head.next;
            //将tmp链表的下一个节点加入小根堆
            if (tmp.next != null) {
                priorityQueue.add(tmp.next);
            }
        }
        return res.next;
    }
}

最小值问题,多去想象小根堆实现,小根堆的维护确保了堆顶一定是最小元素