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.

Analysis:

1. The fast pointer points to the nth node after the head first.

2. Both of the fast and slow pointers start to move at the same speed, which is 1 node per time. 

3. When the fast pointer points to null, the node pointed by the slow pointer should be the one that needs to be deleted. 

/** * 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) {        ListNode newhead = new ListNode(0);        newhead.next = head;        ListNode pslow = newhead;        ListNode slow = head;        ListNode fast = head;                if(head == null) return head;        for(int i=1; i<=n; i++) {            fast = fast.next;        }        while(fast != null) {            pslow = slow;            slow = slow.next;            fast = fast.next;        }        pslow.next = slow.next;                return newhead.next;    }}

Update 02/03/2014: 

/** * 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 || n==0) return null;                ListNode newhead = new ListNode(-1);        newhead.next = head;        ListNode fast = head;        ListNode slow = newhead;    // slow points to one element before the one which should be deleted!!!                for(int i=1; i<=n; i++) {            if(fast!=null) fast=fast.next;            else return null;        }        while(fast != null) {            fast = fast.next;            slow = slow.next;        }        slow.next = slow.next.next;                return newhead.next;    }}