Leetcode: 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.

Understand the problem:
This is a classical two-pointer problem of linked list. One thing needs to handle is deleting the head node. 

Solution:

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        if (head == null || head.next == null) return null;
        
        ListNode slow = head;
        ListNode fast = head;
        
        // move fast pointer n steps ahead of slow
        for (int i = 0; i < n; i++) {
            fast = fast.next;
        }
        
        // move slow and fast pointer one step a time
        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next;
        }
        
        // delete the node
        if (fast == null) {
            head = head.next;
        } else {
            slow.next = slow.next.next;
        }
        return head;
    }
}

Summary:
This is an easy question. However, when deal with linked list problem, make sure you cover all the corner cases.