如何使用Python实现链表的删除链表的倒数第n个节点 II 操作

class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

class Solution:
    def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
        dummy = ListNode(0, head)
        slow, fast = dummy, dummy
        for i in range(n):
            fast = fast.next
        while fast.next:
            slow = slow.next
            fast = fast.next
        slow.next = slow.next.next
        return dummy.next

# test
node1 = ListNode(1)
node2 = ListNode(2)
node3 = ListNode(3)
node4 = ListNode(4)
node5 = ListNode(5)
node1.next = node2
node2.next = node3
node3.next = node4
node4.next = node5

solution = Solution()
new_head = solution.removeNthFromEnd(node1, 2)
while new_head:
    print(new_head.val, end=' ')
    new_head = new_head.next
# Output: 1 2 3 5