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.

/** * Created by ustc-lezg on 16/4/6. */public class Solution {    public ListNode removeNthFromEnd(ListNode head, int n) {        ListNode curr = head;        int len = 0;        while (curr != null) {            curr = curr.next;            ++len;//计算链表的长度        }        if (len == n) {//去掉第一个节点            return head.next;        }        int pos = len - n - 1;//要去掉节点的前一个节点下标        curr = head;        while (pos > 0) {            curr = curr.next;            --pos;        }        curr.next = curr.next.next;        return head;    }}