連結串列中的節點個數大於等於n

給出連結串列1->2->3->4->5->null和 n = 2.

刪除倒數第二個節點之後，這個連結串列將變成1->2->3->5->null

 ListNode removeNthFromEnd(ListNode head, int n) {
        // write your code here
        ListNode dummy = new ListNode(-1);
        dummy.next = head;
        ListNode fast = dummy;
        ListNode slow = dummy;
        while (fast != null && n > -1) {
            fast = fast.next;
            n--;
        }
        while (fast != null) {
            fast = fast.next;
            slow = slow.next;
        }
        slow.next = slow.next.next;
        return dummy.next;
    }