题解 | #单链表的排序#

import java.util.*;

/*
 * public class ListNode {
 *   int val;
 *   ListNode next = null;
 * }
 */

// 空间复杂度nlogn就用归并排序，归并分为两部分，拆分和合并
public class Solution {
    /**
     * 
     * @param head ListNode类 the head node
     * @return ListNode类
     */
    public ListNode sortInList (ListNode head) {

        // 首先判断是否为空或一位
        if(head == null || head.next == null){
            return head;
        }

        // 拆分（利用双指针在中心点拆分,最后slow所在位置就是中点,断开）
        ListNode slow = head;
        ListNode fast = head.next;
        while(fast != null && fast.next != null){
            slow = slow.next;
            fast = fast.next.next;
        }
        ListNode right = slow.next; // 这里用一个right记录右侧的head
        slow.next = null;

        return mergeSort(sortInList(head),sortInList(right));
    }


    // 归并排序
    public ListNode mergeSort(ListNode head1,ListNode head2){

        // 直接用一个ListNode引出一条新的链表记录数据
        ListNode head = new ListNode(-999);
        ListNode followHead = head;

        while(head1 != null && head2 != null){
            if(head1.val > head2.val){
                followHead.next = head2;
                head2 = head2.next;
            }else{
                followHead.next = head1;
                head1 = head1.next;
            }
            followHead = followHead.next;
        }

        followHead.next = head1==null ? head2 : head1;

        return head.next;

    }
}