# Remove Nth Node From End of List

```
Given a linked list, remove the nth 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.
```

分析

两个指针，一个先前进n，然后另一个一起移动，直到第一个达到尾部。

**这题非要2个指针都从dummy起，从Head起就是不行，虽然做法完全一样。**

**因为可能删除的是头结点，所以一定要从dummy起。**

```
/**
 * 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;
        ListNode quick = dummy, slow = dummy;
        for(int i = 0; i <= n; i++){
            quick = quick.next;
        }

        while(quick != null){           
            slow = slow.next;
            quick = quick.next;
        }

            slow.next = slow.next.next;

        return dummy.next;
    }
}
```
