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
        return sort(head, null);
    }
    private ListNode sort(ListNode head, ListNode tail){
      // 递归出口：链表元素个数小于等于1个  
	  if(head == null)return head;
        if(head.next == tail){
            head.next = null;
            return head;
        }
	  // 找到中间节点是关键
        ListNode slow = head, fast = head;
        while(fast != tail){
            slow = slow.next;
            fast = fast.next;
            if(fast != tail)fast = fast.next;
        }
        return merge(sort(head, slow), sort(slow, tail));
    }
    private ListNode merge(ListNode l1, ListNode l2){
        ListNode dummy = new ListNode(-1);
        ListNode cur = dummy;
        while(l1 != null && l2 != null){
            if(l1.val < l2.val){
                cur.next = l1;
                l1 = l1.next;
            }else{
                cur.next = l2;
                l2 = l2.next;
            }
            cur = cur.next;
        }
        cur.next = l1 == null ? l2 : l1;
        return dummy.next;
    }
}

自顶向下的归并排序，时间复杂度可以满足O(nlogn)