[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) {}
 * };
 */
class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        ListNode dummy(0);
        dummy.next = head;
        ListNode *prev = &dummy;
        ListNode *nthNode = head;
        while(head)
        {
            if(0 == n)
            {
                prev = nthNode;
                nthNode = nthNode->next;
            }
            else --n;
            head = head->next;
        }
        
        prev->next = nthNode->next;
        delete nthNode;
        return dummy.next;
    }
};