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.

Use a faster node to move n steps, then use a slower pointer to move along with the faster node, when the faster node is null, the slower node reaches the to be deleted node. Need to check if the to be deleted one is the first node, if it is return head.next.

public ListNode removeNthFromEnd(ListNode head, int n) {
        if (head==null) {
            return null;
        }

        ListNode node = head;

        while (n>0) {
            node = node.next;
            n--;
        }

        ListNode newHead = head;
        ListNode previous = null;
        //move node to end.next of the list
        while (node!=null) {
            previous = head;
            head = head.next;
            node = node.next;            
        }

        //if the deleted one is the first element in the list
        if (previous==null) {
            newHead = head.next;
        } else {
            //set previous node to the one after the deleted node
            previous.next = head.next;
        }    
        return newHead;        
    }