算法描述：

Given a linked list, remove the n-th node from the end of list and return its head.

Example:

Given linked list: 1->2->3->4->5, and n = 2.

After removing the second node from the end, the linked list becomes 1->2->3->5.

Note:

Given n will always be valid.

Follow up:

Could you do this in one pass?

解題思路：

這個題考查數據結構。設置兩個指針，快指針先走n步，然後快指針和慢指針一起走，快指針停的時候，慢指針的下一個元素為待刪除元素。

    ListNode* removeNthFromEnd(ListNode* head, int n) {
        ListNode* dup = new ListNode(-1);
        dup->next = head;
        ListNode* fast = dup;
        while(n>0){
            fast=fast->next;
            n
 
--;
        }
        ListNode* slow = dup;
        while(fast!=nullptr && fast->next!=nullptr){
            fast=fast->next;
            slow=slow->next;
        }
        slow->next = slow->next->next;
        return dup->next;
    }

LeetCode-19-Remove Nth Node From End of List