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 }