链表翻转，确定链表是否成环

链表创建遍历插入删除
reverse1(node * head)
{
        node*p,*q,*r;
        if( head != NULL )
        {
                p = head;
                q = head->next;
                head->next = NULL;
                while(q){
                        r = q->next;
                        q->next = p;
                        p = q;
                        q = r;
                }
                head=p;
        }
        return head;
}

链表是否成环：龟兔赛跑，兔跳两格，龟爬一格，兔子为空，必不成环，相遇则成环。
while(fastPointer != null && fastPointer.next != null)
{ 
    slowPointer = slowPointer.next; 
    fastPointer = fastPointer.next.next; 
    //两指针相遇则有环 
    if(slowPointer == fastPointer)
    { 
        return true; 
    }
}