Leetcode Remove Nth Node From End of List problem solution

class Solution:
    def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:        
        slow = fast = head
        
        for i in range(n):
            fast = fast.next 
            
        if not fast:
            return head.next 
        
        while fast and fast.next:
            slow = slow.next 
            fast = fast.next 
        
        slow.next = slow.next.next 
        return head

public ListNode removeNthFromEnd(ListNode head, int n) {
    if( head == null )
        return null;
    ListNode fakeHead = new ListNode(-1);
    fakeHead.next = head;
    
    ListNode prev = fakeHead, slow = head;
    ListNode fast = head;
    for(int i = 1; i < n; i++)
        fast = fast.next;
    while(fast.next != null) {
        fast = fast.next;
        prev = slow;
        slow = slow.next;
    }
    prev.next = slow.next;
    return fakeHead.next;
}

class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {

        //checking basic  conditions
        if(head==NULL)return head;
        if(n==0)return head;

        ListNode* slow = head;
        ListNode* fast = head;
        ListNode* prev = NULL;

        while(n!=0){
            fast = fast->next;
            n--;
        }

        while(fast!=NULL){
            prev = slow;
            fast=fast->next;
            slow=slow->next;
        }

        if(prev==NULL){
            ListNode* newHead = slow->next;
            delete slow;
            return newHead;
        }

        prev->next = slow->next;
        delete slow;
        return head;
    }
};

struct ListNode* removeNthFromEnd(struct ListNode* head, int n) 
{
    struct ListNode* current;
    current = head;
    int count = 0;
    
    while(current != NULL)
    {
        current = current->next;
        count++;
    }
    
    current = head;
    if(count==n)
        return current->next;
    
    count = count-n-1;
    while(count--)
    {
        current = current->next;
    }
    
    current->next = current->next->next;
    
    return head;
}