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.

Solution:
1. create a dummy head, whose next node is the head to save the head.
2. use two pointers, slow and fast, fast is n steps further than slow. when fast reaches the end of the list, slow should be right before the nth node from the end. remove that node.

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
 // always think about runner is linked list questions
public class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        if(n == 0) return null;
        ListNode tempHead = new ListNode(0);  //used for special case, like: {1}, 1 or n > nodeNum; 
        tempHead.next = head;
        ListNode fast = tempHead;
        ListNode slow = tempHead;
        while(n > 0){
            fast = fast.next;
            n--;
        } 
        while(fast!=null && fast.next!=null){
            slow = slow.next;
            fast = fast.next;
        }
        slow.next = slow.next.next;
        return tempHead.next;
    }
}