Q143 重排链表

给定一个单链表 L：L0→L1→…→Ln-1→Ln ，
将其重新排列后变为： L0→Ln→L1→Ln-1→L2→Ln-2→…

你不能只是单纯的改变节点内部的值，而是需要实际的进行节点交换。

示例 1:

给定链表 1->2->3->4, 重新排列为 1->4->2->3.

示例 2:

给定链表 1->2->3->4->5, 重新排列为 1->5->2->4->3.

/**
* Definition for singly-linked list.
* public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
* }
*/
class Solution {
    public void reorderList(ListNode head) {
        if (head == null || head.next == null || head.next.next == null)
            return;

        ListNode node1 = head;
        ListNode node2 = head;

        while (node2.next != null && node2.next.next != null) {
            node2 = node2.next.next;
            node1 = node1.next;
        }

        ListNode tail = node2.next == null ? node2 : node2.next;

        node2 = node1.next;
        node1.next = null;
        node1 = node2;
        node2 = node1.next;
        node1.next = null;
        while (node2 != null) {
            ListNode node3 = node2.next;
            node2.next = node1;
            node1 = node2;
            node2 = node3;
        }

        node1 = head;
        node2 = tail;

        while (node2 != null) {
            ListNode temp = node2;
            node2 = node2.next;
            temp.next = node1.next;
            node1.next = temp;
            node1 = temp.next;
        }
    }
}