算法题LC8：reorder-list

链表：
题目描述：
将给定的单链表L： L 0→L 1→…→L n-1→L n,
重新排序为： L 0→L n →L 1→L n-1→L 2→L n-2→…
要求使用原地算法，并且不改变节点的值
例如：
对于给定的单链表{1,2,3,4}，将其重新排序为{1,4,2,3}.
Given a singly linked list L: L 0→L 1→…→L n-1→L n,
reorder it to: L 0→L n →L 1→L n-1→L 2→L n-2→…
You must do this in-place without altering the nodes’ values.

For example,
Given{1,2,3,4}, reorder it to{1,4,2,3}.
输入描述：

输出描述：

示例1：
输入：

输出：
        

代码：

public class Solution {
    public void reorderList(ListNode head) {
        if(head == null || head.next == null)
            return;
         
        // 快满指针找到中间节点
        ListNode fast = head;
        ListNode slow = head;
        while(fast.next != null && fast.next.next != null){
            fast = fast.next.next;
            slow = slow.next;
        }
        // 拆分链表，并反转中间节点之后的链表
        ListNode after = slow.next;
        slow.next = null;
        ListNode pre = null;
        while(after != null){
            ListNode temp = after.next;
            after.next = pre;
            pre = after;
            after = temp;
        }
        // 合并两个链表
        ListNode first = head;
        after = pre;
        while(first != null && after != null){
            ListNode ftemp = first.next;
            ListNode aftemp = after.next;
            first.next = after;
            first = ftemp;
            after.next = first;        
            after = aftemp;        
        }
    }
}