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?

/**
 * 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) {
        if(head == null) return null;
        ListNode fast = head;
        while(n-- > 0) {
            fast = fast.next;
        }
        if(fast == null) return head.next;
        ListNode slow = head;
        while(fast.next != null) {
            slow = slow.next;
            fast = fast.next;
        }
        slow.next = slow.next.next;
        return head;
    }
}

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