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.

Define a runner node that runs n steps ahead of head. Walk the head and the node at the same pace, until runner comes to null, delete head.

/** * 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) {        if (head == null) {            return null;        }                ListNode runner = head;        while (n > 0) {            runner = runner.next;            n--;        }                ListNode dummy = new ListNode(0);        dummy.next = head;        head = dummy;                while (runner != null) {            head = head.next;            runner = runner.next;        }        head.next = head.next.next;                return dummy.next;    }}