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.

 

 这道题要注意的Corner Case是：如果n比这个LinkedList的size大，那么就需要直接返回Head，如果相等，就把head删掉，返回head.next；基本做的方法呢，还是Runner Technique. 由于有可能head会被删掉，所以最好还是使用一个dummy node，dummy.next = head。

 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         if (head == null) return null;

15         if (n <= 0) return head;

16         ListNode dummy = new ListNode(-1);

17         dummy.next = head;

18         ListNode runner = dummy;

19         ListNode walker = dummy;

20         int i = 1;

21         while (i<=n && runner.next!=null) {

22             i++;

23             runner = runner.next;

24         }

25         if (i <= n) return head; // n > list's size

26         while (runner.next != null) {

27             runner = runner.next;

28             walker = walker.next;

29         }

30         walker.next = walker.next.next;

31         return dummy.next;

32     }

33 }