题解 | #单链表的排序# 归并排序，只使用递归栈空间

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 head ListNode类 the head node
     * @return ListNode类
     */
    public ListNode sortInList (ListNode head) {
        // 终止条件，递归出口
        if (head == null || head.next == null) {
            return head;
        }

        // 快慢指针找到链表中点，慢指针是链表第二段的首节点
        ListNode slow = head;
        ListNode tail = slow;
        ListNode fast = head;
        while (fast != null) {
            tail = slow;
            slow = slow.next;
            fast = fast.next;
            if (fast != null) {
                fast = fast.next;
            }
        }

        // 很关键，把链表分成两段，上一段指向null
        tail.next = null;

        // 分别排序两部分链表
        ListNode sortedLeft = sortInList(head);
        ListNode sortedRight = sortInList(slow);
        
        // 合并排序链表
        ListNode hair = new ListNode(0);
        ListNode pointer = hair;
        while (sortedLeft != null && sortedRight != null) {
            if (sortedLeft.val < sortedRight.val) {
                pointer.next = sortedLeft;
                sortedLeft = sortedLeft.next;
                pointer = pointer.next;
            } else {
                pointer.next = sortedRight;
                sortedRight = sortedRight.next;
                pointer = pointer.next;
            }
        }
        if (sortedLeft != null) {
            pointer.next = sortedLeft;
        }
        if (sortedRight != null) {
            pointer.next = sortedRight;
        }
        return hair.next;
    }
}