# Remove Nth Node From End of List

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?

## Solution

Count length of the list then subtract N and when iterate through the list up to that new length.

```
public ListNode removeNthFromEnd(ListNode head, int n) {
    ListNode dummy = new ListNode(0);
    dummy.next = head;
    int length  = 0;
    ListNode first = head;
    while (first != null) {
        length++;
        first = first.next;
    }
    length -= n;
    first = dummy;
    while (length > 0) {
        length--;
        first = first.next;
    }
    first.next = first.next.next;
    return dummy.next;
}
```

Find first "slow" element. Then iterate from the first one and update the second one. Then move remove a node.

```
public ListNode removeNthFromEnd(ListNode head, int n) {
    ListNode dummy = new ListNode(0);
    dummy.next = head;
    ListNode first = dummy;
    ListNode second = dummy;
    // Advances first pointer so that the gap between first and second is n nodes apart
    for (int i = 1; i <= n + 1; i++) {
        first = first.next;
    }
    // Move first to the end, maintaining the gap
    while (first != null) {
        first = first.next;
        second = second.next;
    }
    second.next = second.next.next;
    return dummy.next;
}
```