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.

 

各种情况没想全，重复提交了很多次，需要再更加仔细的思考……

/**
 * 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) {
        int i = 0;
            ListNode *p = head;

            for(i = 0; i < n-1 ; i++){      //notice second node from the end just one step from the end
                    p = p->next;
            }
            ListNode *r = NULL;             //r is the poiter ahead of q
            ListNode *q = head;
            while(p->next != NULL){
                p = p->next;
                r = q;
                q = q->next;
            }
            if(r == NULL)
                 head = head->next;
            else{

                r->next = q->next;
            }
            return head;
        }
    
};