[LeetCode]Remove N-th Node from the end of the 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.

class Solution {
public:
ListNode *removeNthFromEnd(ListNode *head, int n) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
if(n<1) return head;
if(head==NULL) return head;
if(head->next==NULL && n==1) return head->next;
if(head->next==NULL && n>1) return head;

ListNode *f1=head; //to locate the previous node of the deleted one
ListNode *f2=head->next;//to locate the node to be deleted
ListNode *f3=head;

int i=1;
while(f3!=NULL && i<=n+1)
{
f3=f3->next;
i++;
}//now f3 points to the next n-th node of the head

if(i<n+1) //n is larger than the list size
return head;
if(i==n+1 && f3==NULL) //delete the head
return head->next;

while(f3!=NULL)
{
f3=f3->next;
f1=f1->next;
f2=f2->next;
}
if(f2==NULL)
f1->next=f2;
else
f1->next=f2->next;
return head;
}
};