# [Remove Nth Node From End of List][title]

## Description

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?

**Tags:** Linked List, Two Pointers


## 思路

题意是让你删除链表中的倒数第 n 个数，我的解法是利用双指针，这两个指针相差 n 个元素，当后面的指针扫到链表末尾的时候，自然它前面的那个指针所指向的下一个元素就是要删除的元素，即 `pre.next = pre.next.next;`，但是如果一开始后面的指针指向的为空，此时代表的意思就是要删除第一个元素，即 `head = head.next;`。

```java
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        ListNode pre = head;
        ListNode afterPreN = head;
        while (n-- != 0) {
            afterPreN = afterPreN.next;
        }
        if (afterPreN != null) {
            while (afterPreN.next != null) {
                pre = pre.next;
                afterPreN = afterPreN.next;
            }
            pre.next = pre.next.next;
        } else {
            head = head.next;
        }
        return head;
    }
}
```


## 结语

如果你同我一样热爱数据结构、算法、LeetCode，可以关注我 GitHub 上的 LeetCode 题解：[awesome-java-leetcode][ajl]



[title]: https://leetcode.com/problems/remove-nth-node-from-end-of-list
[ajl]: https://github.com/Blankj/awesome-java-leetcode