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。

 1 /**
 2  * Definition for singly-linked list.
 3  * public class ListNode {
 4  *     int val;
 5  *     ListNode next;
 6  *     ListNode(int x) {
 7  *         val = x;
 8  *         next = null;
 9  *     }
10  * }
11  */
12 public class Solution {
13     public ListNode removeNthFromEnd(ListNode head, int n) {
14         // Note: The Solution object is instantiated only once and is reused by each test case.
15         ListNode per = head;
16         ListNode cur = head;
17         ListNode before = head;
18         int len = 0;
19         while(per != null){
20             per = per.next;
21             len ++;
22         }
23         int num = len - n;
24         if(num == 0)return head.next;
25         else{
26             per = head;
27             for(int i = 0; i < num; i ++){
28                 cur = per;
29                 per = per.next;
30             }
31             cur.next = per.next;
32             return head;
33         }
34     }
35 }

第二遍：

 1 public class Solution {
 2     public ListNode removeNthFromEnd(ListNode head, int n) {
 3         // Note: The Solution object is instantiated only once and is reused by each test case.
 4         ListNode header = new ListNode(-1);
 5         header.next = head;
 6         ListNode cur = header;
 7         ListNode per = header;
 8         ListNode fast = header;
 9         for(int i = 0; i < n; i ++){
10             fast = fast.next;
11         }
12         while(fast != null){
13             per = cur;
14             fast = fast.next;
15             cur = cur.next;
16         }
17         per.next = cur.next;
18         return header.next;
19     }
20 }