Remove Nth Node From End of List

PROBLEM :

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.

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SIMPLE c++ IMPLEMENTATION :
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/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* removeNthFromEnd(struct ListNode* head, int n) {
   
    struct ListNode *n1,*n2,*temp ;
    int i=1 ;
   
    n1=head ;
    n2=head ;
   
    while(i!=n)
    {
        n1=n1->next ;
        i++ ;
    }
    if(n1->next==NULL)
    {
        temp=head ;
        head=head->next ;
        free(temp) ;
       
        return head ;
    }
   
   
    i=1 ;
    n1=head ;
    n2=head->next ;
   
    while(i<n&&i!=n)
    {
        n2=n2->next ;
        i++ ;
    }
   
    while(n2->next!=NULL)
    {
        n1=n1->next ;
        n2=n2->next ;
    }
   
    temp=n1->next ;
    n1->next=temp->next ;
    free(temp) ;
   
    return head ;
}

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