前往此题

辅助栈，模拟法

1. 将链表节点放入栈stack中
2. 再创建新的链表，按照1->stack.pop()->2->stack.pop()->3 这样的形式进行拼接即可

# 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.
        """
        stack = []
        cur = head
        while cur:
            stack.append(cur)
            cur = cur.next
        cur = head
        res = ListNode(0)

        while cur.next != res.next:
            res = stack.pop()
            res.next = cur.next
            cur.next = res
            cur = cur.next.next
        res.next = None