題目描述：

方法一：這是我最開始想到的方法，簡答直接。先找到連結串列的長度count，然後刪除第count-n+1個節點。需要兩遍遍歷。複雜度為o(n),n為連結串列的長度。
程式碼：

class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        ListNode* find = head;
        long count = 1;

        while (find->next != NULL)
        {
            ++count;
            find = find->next
 
;
        }

        ListNode *curr = head;
        ListNode *pre = NULL;

        for (int i = 1; i <= count - n; ++i)
        {
            pre = curr;
            curr = curr->next;
        }

        if (curr == head)
            head = curr->next;
        else 
        pre->next = curr->next
 
;

       delete curr;

        return head;

    }
};

方法二：方法一的複雜度為o(n),已經是最優的複雜度，因為連結串列的操作只能通過頭一步步到達某個節點，就好像逐個比較法一樣。但是，還可以改進，使用兩個指標同時遍歷，保持前後指標距離為n個節點，遍歷一次，就可完成任務，使計算量大概下降一半。
程式碼：

class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        ListNode *first = head;
        ListNode *second
 
 = head;
        long count = 0;

        while (first->next != NULL)
        {
            ++count;
            first = first->next;

            if (count > n)
                second = second->next;
        }

        if (count < n)
        {
             head = head->next;
            delete second;
        }

        else
        {
            first = second->next;
            second->next = second->next->next;
            delete first;

        }


        return head;

    }
};