题解 | #单链表的排序#

import java.util.*;

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

public class Solution {
    /**
     *
     * @param head ListNode类 the head node
     * @return ListNode类
     */
    public ListNode sortInList (ListNode head) {
        // write code here
        if(head == null) return null;
        if(head.next == null) return head; //排除空链表 和只有一个Node的链表的情况 
        ListNode pre = head;
        ListNode slow = head;
        ListNode fast = head;
        while(fast!=null&&fast.next!=null){
            pre = slow;
            slow = slow.next;
            fast = fast.next.next;
        }
        pre.next = null; //将链表从中间一分为二
        ListNode head1 = sortInList(head);
        ListNode head2 = sortInList(slow);
        return mergeSortList(head1,head2);
    }
    public ListNode mergeSortList(ListNode list1, ListNode list2) {
        ListNode dummy = new ListNode(0);
        ListNode cur = dummy;
        while (list1 != null && list2 != null) {
            if (list1.val <= list2.val) {
                cur.next = list1;
                list1 = list1.next;
            } else {
                cur.next = list2;
                list2 = list2.next;
            }
            cur = cur.next;
        }
        if(list1 == null){
            cur.next = list2;
        }
        if(list2 == null){
            cur.next = list1;
        }
        return dummy.next;
    }
}