思路：分别处理奇偶，保存奇偶的第一个和最后一个节点，注意最后链接的时候需要把偶数的next去掉再拼接不然就成环了

class Solution:
    def oddEvenList(self, head: ListNode) -> ListNode:
        if not head or not head.next or not head.next.next:
            return head

        first_odd = head
        first_even = head.next
        last_odd = head
        last_even = head.next
        current_node = first_even.next
        node_num = 3

        while current_node:
            next_node = current_node.next
            if node_num % 2 == 1:
                last_odd.next = current_node
                last_odd=current_node
            else:
                last_even.next = current_node
                last_even=current_node
            current_node = next_node
            node_num += 1
        last_odd.next=first_even
        last_even.next=None
        return first_odd