算法思想： 先分析是否有环。为此我们建立两个指针，从header一起向前跑，一个步长为1，一个步长为2，用while（直到步长2的跑到结尾）检查两个指针是否相等，直到找到为止。
static bool JudgeCircleExists(Link head){Link first = head; //1 step each timeLink second = head; //2 steps each timewhile (second.Next != null && second.Next.Next != null){second = second.Next.Next;first = first.Next;if (second == first)return true;}return false;}那又如何知道环的长度呢? 根据上面的算法，在返回true的地方，也就是2个指针相遇处，这个位置的节点P肯定位于环上。我们从这个节点开始先前走，转了一圈肯定能回来：
static int GetCircleLength(Link point){int length = 1;Link curr = point;while (curr.Next != point){length++;curr = curr.Next;}return length;}继续我们的讨论，如何找到环的“起始”点呢? 延续上面的思路，我们仍然在返回true的地方P，计算一下从有环单链表的表头head到P点的距离。
static int GetLengthFromHeadToPoint(Link head, Link point){int length = 1;Link curr = head;while (curr != point){length++;curr = curr.Next;}return length;}如果我们把环从P点“切开”（当然并不是真的切，那就破坏原来的数据结构了），那么问题就转化为计算两个相交“单链表”的交点。一个单链表是从P点出发，到达P（一个回圈），距离M；另一个单链表从有环单链表的表头head出发，到达P，距离N。
PRivate static Link FindIntersect(Link head){Link p = null;//get the point in the circlebool result = JudgeCircleExists(head, ref p);if (!result) return null;Link curr1 = head.Next;Link curr2 = p.Next;//length from head to pint M = 1;while (curr1 != p){M++;curr1 = curr1.Next;}//circle lengthint N = 1;while (curr2 != p){N++;curr2 = curr2.Next;}//recover curr1 & curr2curr1 = head.Next;curr2 = p.Next;//make 2 links have the same distance to the intersectif (M > N){for (int i = 0; i < M – N; i++)curr1 = curr1.Next;}else if (M < N){for (int i = 0; i < N – M; i++)curr2 = curr2.Next;}//goto the intersectwhile (curr1 != p){if (curr1 == curr2){return curr1;}curr1 = curr1.Next;curr2 = curr2.Next;}return null;}新闻热点
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