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 time
Link second = head; //2 steps each time
while (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（第10题）：
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。
我们可以参考第10题的GetIntersect方法并稍作修改。
private static Link FindIntersect(Link head)
{
Link p = null;
//get the point in the circle
bool result = JudgeCircleExists(head, ref p);
if (!result) return null;
Link curr1 = head.Next;
Link curr2 = p.Next;
//length from head to p
int M = 1;
while (curr1 != p)
{
M++;
curr1 = curr1.Next;
}
//circle length
int N = 1;
while (curr2 != p)
{
N++;
curr2 = curr2.Next;
}
//recover curr1 & curr2
curr1 = head.Next;
curr2 = p.Next;
//make 2 links have the same distance to the intersect
if (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 intersect
while (curr1 != p)
{
if (curr1 == curr2)
{
return curr1;
}
curr1 = curr1.Next;
curr2 = curr2.Next;
}
return null;
}