题解 | #单链表的排序#

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 head ListNode类 the head node
     * @return ListNode类
     */
    public ListNode sortInList (ListNode head) {
        // write code here
        if(head==null||head.next==null){
            return head;
        }
        return sort(head);
    }

    private ListNode sort(ListNode head) {
        // TODO
        if(head==null||head.next==null){
            return head;
        }
        ListNode middle = getMiddle(head);
        head = sort(head);
        middle = sort(middle);
        return merge(head,middle);
    }

    private ListNode merge(ListNode head, ListNode middle) {
        // TODO
        ListNode res = new ListNode(Integer.MIN_VALUE);
        ListNode cur = res;
        ListNode cur1 = head;
        ListNode cur2 = middle;
        while(cur1!=null&&cur2!=null){
            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 = cur1;
        }else{
            cur.next = cur2;
        }
        return res.next;
    }

    private ListNode getMiddle(ListNode head) {
        // TODO
        ListNode fast = head.next;
        ListNode slow = head;
        while(fast!=null&&fast.next!=null){
            fast = fast.next.next;
            slow = slow.next;
            
        }
        ListNode middle = slow.next;
        slow.next = null;
        return middle;

    }
}