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) {
if(head==NULL) return NULL;
ListNode* fast=head;
ListNode* slow=head;
ListNode* prev=NULL;
for(int i=1;i<=n;i++){
if(fast==NULL) return NULL;
fast=fast->next;
}
//if(fast==NULL) return head;
while(fast){
fast=fast->next;
prev=slow;
slow=slow->next;
}
if(prev!=NULL && slow!=head)
prev->next=slow->next;
else head=head->next;
return head;
}
};