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.

 /**  
  * Definition for singly-linked list.  
  * struct ListNode {  
  *   int val;  
  *   ListNode *next;  
  *   ListNode(int x) : val(x), next(NULL) {}  
  * };  
  */  
   typedef ListNode ln;  
   
   ListNode *removeNthFromEnd(ListNode *head, int n) {  
     ln *dummy = new ln(0);  
     dummy->next = head;  
     ln *fast = dummy, *slow = dummy;  
     while(n--)  
     {  
       fast = fast->next;  
     }  
       
     while(fast->next)  
     {  
       fast = fast->next;  
       slow = slow->next;  
     }  
       
     slow->next = slow->next->next;  
       
     return dummy->next;  
   }