题解 |归并排序 #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
     */

     /**
     合并有序链表
      */
    public ListNode merge(ListNode p1, ListNode p2){
        ListNode dummy = new ListNode(-1);
        ListNode cur = dummy;
        while(p1 != null && p2 != null){
            if(p1.val <= p2.val){
                cur.next = p1;
                p1 = p1.next;
            }else{
                cur.next = p2;  
                p2 = p2.next;        
            }
            cur = cur.next;  
        }
        if(p1 != null){
            cur.next = p1;
        }
        if(p2 != null){
            cur.next = p2;
        }
        return dummy.next;
    }


    public ListNode sortInList (ListNode head) {
        // write code here
        if(head == null || head.next == null){
            return head;
        }
        //一：分割环节。使用快慢指针寻找链表中点，如果是偶数，则fast走到终点时，slow位于中间偏左
        ListNode fast = head.next;
        ListNode slow = head;
        while(fast != null && fast.next!=null){
            fast = fast.next.next;
            slow = slow.next;
        }
        //在中点处将链表一分为二
        ListNode tmp = slow.next;
        slow.next = null;
        //二：合并环节。递归进行左右两边的排序
        ListNode left = sortInList(head);
        ListNode right = sortInList(tmp);
        return merge(left,right);
    }
}