# 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。 
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#  進階：你能嘗試使用一趟掃描實現嗎？ 
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#  示例 1： 
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# 輸入：head = [1,2,3,4,5], n = 2
# 輸出：[1,2,3,5]

方法：雙指標

# leetcode submit region begin(Prohibit modification and deletion)
# 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:
        dummy = ListNode(0, head)   # 啞節點指向頭結點
        first = head                # 快指標指向頭結點
        second = dummy              # 慢指標指向啞節點
        # 快指標先走n
        for
 
 i in range(n):
            first = first.next
        # 直至快指標為空，到連結串列結尾
        while first:
            first = first.next
            second = second.next
        # 慢指標的next指向刪除節點的next節點
        second.next = second.next.next
        # 返回頭結點
        return dummy.next
# leetcode submit region end(Prohibit modification and deletion)