Given a linked list, remove the n-th node from the end of list and return its head.

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.

//还是走的快的点(fastNode)与走得慢的点(slowNode)路程差的问题
    public static ListNode removeNthFromEnd(ListNode head, int n) {
        ListNode headNode = new ListNode(9527);
        headNode.next = head;
        ListNode fastNode = headNode;
        ListNode slowNode = headNode;
        while(fastNode.next != null){
            if(n <= 0)
                slowNode = slowNode.next;
            fastNode = fastNode.next;
            n--;
        }
        if(slowNode.next != null)
            slowNode.next = slowNode.next.next;
        return headNode.next;
    }