# 19.Remove Nth Node from End of List

\#19 Remove Nth Node from End of List 难度:Medium

> 给定一个链表，删除链表的倒数第 n 个节点，并且返回链表的头结点。

```
示例：

给定一个链表: 1->2->3->4->5, 和 n = 2.

当删除了倒数第二个节点后，链表变为 1->2->3->5.
说明：

给定的 n 保证是有效的。

进阶：

你能尝试使用一趟扫描实现吗？
```

双指针，

```
/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        ListNode* root=head;
        ListNode* fast=head;
        while(n--)
        fast=fast->next;
        if(!fast) return root->next;
        while(fast->next)
        {
            root=root->next;
            fast=fast->next;

        }
        root->next=root->next->next;;
        return head;
    }
};
```

> 执行用时 :8 ms, 在所有 C++ 提交中击败了45.95%的用户\
> 内存消耗 :6.5 MB, 在所有 C++ 提交中击败了100.00%的用户


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