Remove Nth Node From End of List

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.

/**
 * 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) {
        List<ListNode> list = new ArrayList<ListNode>();
        ListNode temp = head;
        if(null == head.next)
            return null;
        while(null != temp){
            list.add(temp);
            temp = temp.next;
        }//只要一趟，把所有节点保存起来
        
        if(list.size() - n - 1 >= 0){
            temp = list.get(list.size() - n - 1);
            temp.next = temp.next.next;
            return head;
        }
        head = head.next;
        return head;
    }
}

 ps:依然是双指针思想，两个指针相隔n-1，每次两个指针向后一步，当后面一个指针没有后继了，前面一个指针就是要删除的节点。博客园上面看到的，就没有实现了