重排链表

解题思路：

1. 边界处理
2. 找到中间节点
3. 翻转后半段链表
4. 归并链表

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

        ListNode prev = null;
        ListNode cur = midNext;
        while(cur != null){
            ListNode next = cur.next;
            cur.next = prev;
            prev = cur;
            cur = next;
        }
        ListNode node = head;
        while(node != null && prev != null){
            ListNode n1 = node.next;
            ListNode n2 = prev.next;
            node.next = prev;
            prev.next = n1;
            node = n1;
            prev = n2;
        }

    }

    public ListNode getMid(ListNode head){
        ListNode slow = head;
        ListNode fast = head;
        while(fast.next != null && fast.next.next != null){
            slow = slow.next;
            fast = fast.next.next;
        }
        return slow;
    }
}