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) {
        ListNode vNode = new ListNode(-1);
        ListNode head = vNode;
        ListNode min;

        while ((min = get_min(lists)) != null) {
            head.next = min;
            head = head.next;
        }
        return vNode.next;
    }
    ListNode get_min(List<ListNode> nodes) {
        ListNode min = null;
        int index = 0;
        for (int i = 0; i < nodes.size(); i++) {
            ListNode curr = nodes.get(i);
            if (min == null || (curr != null && min.val > curr.val)) {
                min = nodes.get(i);
                index = i;
            }
        }
        if (min != null) {
            nodes.set(index, min.next);
        }
        return min;
    }
}