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
     */
    // 给链表排序 !
    // 采用归并的思想,合并有序链表!
    public ListNode sortInList (ListNode head) {
        // write code here
        if(head == null) return null;
        // 将链表一分为二
        ListNode mid = getMid(head);
        if(mid == null) return head;
        return merageSort(sortInList(head),sortInList(mid));
    }
    
    // 获取链表的中间节点!
    private ListNode getMid(ListNode head){
        if(head == null) return null;
        ListNode fake_head = new ListNode(-1);
        fake_head.next = head;
        ListNode pre = fake_head;
        ListNode slow = head,fast = head;
        while(fast != null){
            pre = pre.next;
            fast = fast.next;
            slow = slow.next;
            if(fast != null) fast = fast.next;
        }
        pre.next = null;
        return slow;
    }
    
    // 合并两个有序链表
    private ListNode merageSort(ListNode l1,ListNode l2 ){
        if(l1 == null) return l2;
        if(l2 == null) return l1;
        if(l1.val < l2.val){
            l1.next = merageSort(l1.next,l2);
            return l1;
        }
        l2.next = merageSort(l1,l2.next);
        return l2;
    }
    
    
}