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.通过次数47,851 | 提交次数82,512 代码实现
# Definition for singly-linked list. class ListNode: def __init__(self, val=0, next=None): self.val = val self.next = next class Solution: def reorderList(self, head: ListNode) -> None: """ Do not return anything, modify head in-place instead. """ if not head or not head.next: return fast,slow = head, head while fast.next and fast.next.next: slow = slow.next fast = fast.next.next p,right = slow.next,None slow.next = None while p: right, right.next, p = p, right, p.next cur = head while cur and right: cur.next, right.next, cur, right = right, cur.next, cur.next, right.next 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。