题解 | #合并k个已排序的链表# 归并排序，开辟0堆空间

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) {
        return divMerge(lists, 0, lists.size() - 1);
    }

    private ListNode divMerge(ArrayList<ListNode> lists, int left, int right) {
        if (left > right) {
            return null;
        }
        if (left == right) {
            return lists.get(left);
        }
        int mid = left + (right - left) / 2;
        ListNode mergedLeft = divMerge(lists, left, mid);
        ListNode mergedRight = divMerge(lists, mid + 1, right);
        ListNode hair = new ListNode(0);
        ListNode pointer = hair;
        while (mergedLeft != null || mergedRight != null) {
            if (mergedLeft == null) {
                pointer.next = mergedRight;
                return hair.next;
            } else if (mergedRight == null) {
                pointer.next = mergedLeft;
                return hair.next;
            } else if (mergedLeft.val < mergedRight.val) {
                pointer.next = mergedLeft;
                mergedLeft = mergedLeft.next;
            } else {
                pointer.next = mergedRight;
                mergedRight = mergedRight.next;
            }
            pointer = pointer.next;
        }
        return hair.next;
    }
}