給你一個連結串列，刪除連結串列的倒數第n個結點，並且返回連結串列的頭結點。

進階：你能嘗試使用一趟掃描實現嗎？

示例 1：

輸入：head = [1,2,3,4,5], n = 2
輸出：[1,2,3,5]
示例 2：

輸入：head = [1], n = 1
輸出：[]
示例 3：

輸入：head = [1,2], n = 1
輸出：[1]

提示：

連結串列中結點的數目為 sz
1 <= sz <= 30
0 <= Node.val <= 100
1 <= n <= sz

方法1(兩次掃描)

# Definition for singly-linked list.
# 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:
        # 兩趟掃描
        l = []
        while head:
            l.append(head.val)
            head = head.next
        
        new_head 
 
= ListNode(0)
        p = new_head

        for i in range(len(l)):
            if i == len(l) - n:
                continue

            p.next = ListNode(l[i])
            p = p.next
            
        return new_head.next
        

方法2(兩次掃描)

# Definition for singly-linked list.
# 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:
        # 兩趟掃描
        l = self.getLength(head)
        dummy = ListNode(0, head)
        p = dummy
        
        i = 0
        while p:
            if i == l - n:
                p.next = p.next.next # 刪除節點
                break
            i += 1
            p = p.next
        
        return dummy.next
    
    def getLength(self, p):
        l = 0
        while p:
            l += 1
            p = p.next
        return l