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.代码
class Solution { public void reorderList(ListNode head) { if(head == null || head.next == null || head.next.next == null) return; // 1. 寻找中点 ListNode slow = head; ListNode fast = head.next; while(fast != null && fast.next != null){ slow = slow.next; fast = fast.next.next; } // 2. 断开中点，反转后半段链表 ListNode head2 = null; ListNode next = slow.next; slow.next = null; slow = next; while(slow != null){ next = slow.next; slow.next = head2; head2 = slow; slow = next; } // 3. 将head和head2进行拼接 ListNode curr = head; ListNode curr2 = head2; while(curr != null && curr2 != null){ next = curr.next; curr.next = curr2; curr2 = curr2.next; curr.next.next = next; curr = next; } } }涉及到的知识点
