Remove Nth Node From End of List

Given a linked list, remove the n^th 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) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        ListNode* p = head;        ListNode* q = head;        while (p) {            if (n != -1) {                --n;            } else {                q = q->next;            }            p = p->next;        }        if (n == 0) {           head = head->next;           delete q;        } else if (n == -1) {            p = q->next;            if (p != NULL) {               q->next = p->next;               delete p;            }        }        return head;            }};