关于链表的三个常用算法

 //找到环的第一个入口点        static public SinglyLinkedListNode<T> FindLoopPort(SinglyLinkedList<T> list)        {            SinglyLinkedListNode<T> pslow = list.head;            SinglyLinkedListNode<T> pfast = list.head;            //为什么有环的单向链表这样做一定会相交？            while (pfast != null && pfast.next != null)            {                pslow = pslow.next;        // 每次前进一步                pfast = pfast.next.next;  // 每次前进二步                if (pslow == pfast)          // 两个指针相遇，说明存在环                    break;            }            if (pfast == null || pfast.next == null)    // 不存在环                return null;            pslow = list.head;            while (pslow != pfast)            {                pslow = pslow.next;        // 每次前进一步                pfast = pfast.next;        // 每次前进一步            }            return pslow;       // 返回环的入口点        }        //两个无环单向链表是否相交, 若相交则求出第一个相交的节点        static public SinglyLinkedListNode<T> Intersection(SinglyLinkedList<T> list1, SinglyLinkedList<T> list2)        {            int len1 = list1.Count;            int len2 = list2.Count;            int distance=Math.Abs(len1-len2);            SinglyLinkedListNode<T> head1=list1.head;            SinglyLinkedListNode<T> head2=list2.head;            if (len1 < len2)            {                for (int i = 0; i < distance; i++)                    head2 = head2.next;            }            else if (len2 < len1)            {                for (int i = 0; i < distance; i++)                    head1 = head1.next;            }            //指针对齐之后开始确定相交节点            while (head1.next != null && head2.next != null && head1 != head2)            {                head1 = head1.next;                head2 = head2.next;            }            if (head1 != null && head2 != null)                return head1;            else                return null;        }        //单向链表的反转        static public void Reverse(SinglyLinkedList<T> list)        {            SinglyLinkedListNode<T> p, q;            SinglyLinkedListNode<T> temp;            if (list.Count >= 2)            {                p = list.head;                q = p.next;            }            else                return;            while (q != null)            {                temp = q.next;                q.next = p;                p = q;                q = temp;            }            list.head.next = null;            list.head = p;        }