刪除鏈表的倒數第N個節點

我的方法是用一個隊列，保存後N個節點，但是內存消耗比較大，方法如下：

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
        stackList = []
        
        from collections import deque
        stackList = deque(stackList)
        p = head
        while p is not None:
            if len(stackList) <= n:
                stackList.append(p)
            else:
                stackList.popleft()
                stackList.append(p)
            p = p.next
        
        if len(stackList) == n: #這裏是爲了應對出現stackList不滿（即不是n+1）的情況
            stackList.popleft()
            try:
                return stackList[0]
            except:
                return None
        stackList[0].next = stackList[1].next
    
        return head

還用一種雙指針方法：

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
        headNode = ListNode(-1) #在head之前增加節點可以避免出現listNode長度和n一致的情況
        headNode.next = head
        front = headNode
        
        end = front
        
        for _ in range(n): #排頭指針先跑n步，保證和隊尾指針相差n，即隊尾指針的next是要刪除的
            front = front.next
            
        while front.next is not None:
            front = front.next
            end = end.next
        end.next = end.next.next
        
        return headNode.next