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
     */

    private ListNode Merge (ListNode pHead1, ListNode pHead2) {
        // write code here
        if (pHead1 == null) {
            return pHead2;
        } else if (pHead2 == null) {
            return pHead1;
        }

        ListNode dummyNode = new ListNode(-1001);  // 用超出范围的哑节点
        ListNode cur = dummyNode;
        ListNode cur1 = pHead1;
        ListNode cur2 = pHead2;

        while (true) {
            if (cur1.val <= cur2.val) {
                cur.next = cur1;
                cur = cur.next;
                cur1 = cur1.next;
            } else {
                cur.next = cur2;
                cur = cur.next;
                cur2 = cur2.next;
            }
            if (cur1 == null) {
                cur.next = cur2;
                break;
            } else if (cur2 == null) {
                cur.next = cur1;
                break;
            }
        }
        return dummyNode.next;
    }

    public ListNode mergeKLists (ArrayList<ListNode> lists) {
        // write code here
        ListNode dummyNode = new ListNode(-1001);
        ListNode pHead1, pHead2;
        pHead1 = dummyNode;
        for(int i=0;i<lists.size();i++){
            pHead2 = lists.get(i);
            pHead1 = Merge(pHead1, pHead2);
        }

        return dummyNode.next;
    }
    }

在合并两个的基础上，循环k次即可；

注意ArrayList类型的索引方法是list.get(i), 获取大小的方法是list.size()