# [LeetCode 19] Remove Nth Node From End of List

### Question

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.

### Stats

 Frequency 3 Diffficulty 2 Adjusted Difficulty 1 Time to use ----------

Ratings/Color = 1(white) 2(lime) 3(yellow) 4/5(red)

### Analysis

Note the special case: if the head node needs to be removed!

### Solution

The code explains itself. Just don’t forget the special cases.

### My code

``````/**
* public class ListNode {
*     int val;
*     ListNode next;
*     ListNode(int x) {
*         val = x;
*         next = null;
*     }
* }
*/
public class Solution {
public ListNode removeNthFromEnd(ListNode head, int n) {
return null;
}
// important to note that head node can be removed as well!
for (int i = 0; i < n; i++) {
right = right.next;
if (right == null) {
}
}
// advance left and right pointer together
while (right.next != null) {
left = left.next;
right = right.next;
}
// remove the node after left pointer
// again, the below error check is not necessary
if (left.next == null) {
// need to remove the header in this case