题解 | #单链表的排序#

1. fast跳两格：fast = fast.next.next
2. slow跳一格：slow = slow.next
3. 若为偶数个则slow在左边那个
4. 若为奇数个则slow在中间那个
5. 递归调用sortinlist
6. 最后添加未对比的链表部分判断左链表是否为空针对的是奇数个情况。

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 || head.next == null)
            return head;
        // 使用快慢指针寻找链表的中点
        ListNode fast = head.next, slow = head;
        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }
        ListNode tmp = slow.next;
        slow.next = null;
        // 递归左右两边进行排序
        ListNode left = sortInList(head);
        ListNode right = sortInList(tmp);
        // 创建新的链表
        ListNode h = new ListNode(0);
        ListNode res = h;
        // 合并 left right两个链表
        while (left != null && right != null) {
            // left  right链表循环对比
            if (left.val < right.val) {
                h.next = left;
                left = left.next;
            } else {
                h.next = right;
                right = right.next;
            }
            h = h.next;
        }
        // 最后添加未对比的链表部分判断左链表是否为空
        h.next = left != null ? left : right;
        return res.next;
    }
}