题解 | #合并k个已排序的链表# 归并排序

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) {
        if (lists.isEmpty()) {
            return null;
        }
        if (lists.size() == 1) {
            return lists.get(0);
        }
        int size = lists.size();
        int leftSize = size / 2;
        int rightSize = size - leftSize;
        ArrayList<ListNode> leftPart = new ArrayList<>();
        ArrayList<ListNode> rightPart = new ArrayList<>();
        for (int i = 0; i < leftSize; i++) {
            leftPart.add(lists.get(i));
        }
        for (int i = leftSize; i < size; i++) {
            rightPart.add(lists.get(i));
        }
        ListNode leftRes = mergeKLists(leftPart);
        ListNode rightRes = mergeKLists(rightPart);
        ListNode hair = new ListNode(0);
        ListNode pointer = hair;
        while (leftRes != null || rightRes != null) {
            if (leftRes == null) {
                pointer.next = rightRes;
                return hair.next;
            } else if (rightRes == null) {
                pointer.next = leftRes;
                return hair.next;
            } else if (leftRes.val < rightRes.val) {
                pointer.next = leftRes;
                leftRes = leftRes.next;
            } else {
                pointer.next = rightRes;
                rightRes = rightRes.next;
            }
            pointer = pointer.next;
        }
        return hair.next;
    }
}