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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.
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/reorder-list 著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode() {} * ListNode(int val) { this.val = val; } * ListNode(int val, ListNode next) { this.val = val; this.next = next; } * } */class Solution { public void reorderList(ListNode head) { if(head==null || head.next==null) return; ListNode mid = midNode(head); ListNode R = reverse(mid.next); mid.next = null; ListNode p = head,q = R; while(q!=null) { ListNode first = p.next; ListNode second = q.next; p.next = q; q.next = first; p = first; q = second; } } public ListNode midNode(ListNode head) { if(head==null || head.next==null) return head; ListNode fast = head; ListNode slow = head; while(fast!=null &&fast.next!=null) { fast = fast.next.next; slow = slow.next; } return slow; } public ListNode reverse(ListNode head) { ListNode temp = null,pre = null; while(head!=null) { temp = head.next; head.next = pre; pre = head; head = temp; } return pre; } }