Remove Nth Node From End of List

Given a linked list, remove the n^th 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.

 也可以用双指针（fast pointer） 来做，但是要注意 n = len的情况。

/**
 * 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) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        ListNode per = head;
        ListNode cur = head;
        ListNode before = head;
        int len = 0;
        while(per != null){
            per = per.next;
            len ++;
        }
        int num = len - n;
        if(num == 0)return head.next;
        else{
            per = head;
            for(int i = 0; i < num; i ++){
                cur = per;
                per = per.next;
            }
            cur.next = per.next;
            return head;
        }
    }
}

 第二遍：

public class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        ListNode header = new ListNode(-1);
        header.next = head;
        ListNode cur = header;
        ListNode per = header;
        ListNode fast = header;
        for(int i = 0; i < n; i ++){
            fast = fast.next;
        }
        while(fast != null){
            per = cur;
            fast = fast.next;
            cur = cur.next;
        }
        per.next = cur.next;
        return header.next;
    }
}