Remove Nth Node From End of List

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.

class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        if (head == nullptr) return head;

        ListNode dummy(-1);
        dummy.next = head;
        ListNode *fast = &dummy;
        ListNode *slow = &dummy;
        for (int i = 0; i < n && fast->next != nullptr; i++) {
            fast = fast->next;
        }

        while (fast->next != nullptr) {
            fast = fast->next;
            slow = slow->next;
        }
        ListNode *tmp = slow->next;
        slow->next = slow->next->next;
        delete tmp;

        return dummy.next;
    }
};

bug version

Case1:
input [1] n = 1
bug at 
   while (fast->next != nullptr)
Line 19: member access within null pointer of type 'struct ListNode'
slow->next = slow->next->next;    // access nullptr

Case2:
input [1 2] n = 2
when delete head node, the bug will occur

need use Dummy Node to delete head

class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        if (head == nullptr) return head;
        ListNode *fast = head;
        ListNode *slow = head;
        for (int i = 0; i < n && fast != nullptr; i++) {
            fast = fast->next;
        }

        while (fast->next != nullptr) {
            fast = fast->next;
            slow = slow->next;
        }
        ListNode *tmp = slow->next;
        slow->next = slow->next->next;
        delete tmp;

        return head;
    }
};