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.

代码

/*** 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) {double n = 0;if(head==null) return;ListNode cnt = head;while (cnt != null) {//统计节点数n++;cnt = cnt.next;}ListNode r = head;for (int i = 0; i < Math.ceil(n / 2)-1; i++) {//找出中间节点r = r.next;}ListNode cur = r.next, pre = null;r.next=null;//将链表切割成两个while (cur != null) {//将后面部分的链表逆序ListNode temp = cur.next;cur.next = pre;pre = cur;cur = temp;}while (pre != null) {//合并两个链表ListNode np = pre.next, nh = head.next;head.next = pre;pre.next = nh;pre = np;head = nh;}}
}