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) {
int l=2;
int i=2;
ListNode *temp,*tt;
temp=head->next;
if(temp==NULL)
{
if(n==0) return head;
else return NULL;
}
while(temp->next!=NULL)
{
temp=temp->next;
l++;
}
if(n==l)
{
return head->next;
free(head);
}
temp=head;
while(temp->next!=NULL)
{

if(i==l-n+1)
{
tt=temp->next;
temp->next=tt->next;
free(tt);
i++;
continue;
}
temp=temp->next;
i++;
}
return head;
}
};