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.
基础题，考验对链表的理解。

/**
 * 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 ln = null;
				int sum = 0;
				ListNode temp = head;
				while (temp != null){
					sum++;
					temp = temp.next;
				}
				if(sum == 0 || n>sum || n<=0){
					return head;
				}
				ln = head;
				int count = sum - n;
				if(count >0){
					while(--count > 0){
						 ln = ln.next;
						if(ln == null){
							return head;
						}
					}
					
					ln.next = ln.next.next;
					
				}else{
					
					head = head.next;
				}
				
				
				return head;
	    }
}