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;    }};