# 19. Remove Nth Node From End of List

[**19. Remove Nth Node From End of List**](https://leetcode.com/problems/remove-nth-node-from-end-of-list/description/)

Given a linked list, remove the \_n\_th node from the end of list and return its head.

For 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.\
Try to do this in one pass.

&#x20;**My Solutions:**

* head先走n步，接着node和head同时走，保持n的距离
* 当head走到末尾时，node就在倒数第n个。

```
/**
 * 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 dummy = new ListNode(0);
        dummy.next = head;

        for (int i = 0; i < n; i++) head = head.next;

        ListNode node = dummy;

        while (head != null) {
            head = head.next;
            node = node.next;
        }
        node.next = node.next.next;
        return dummy.next;
    }
}
```
