这道题有多种算法。 算法1：把第一个链表逐项存在hashtable中，遍历第2个链表的每一项，如果能在第一个链表中找到，则必然相交。
static bool JudgeIntersectLink1(Link head1, Link head2){Hashtable ht = new Hashtable();Link curr1 = head1;Link curr2 = head2;//store all the elements of link1while (curr1.Next != null){ht[curr1.Next] = string.Empty;curr1 = curr1.Next;}//check all the elements in link2 if exists in Hashtable or notwhile (curr2.Next != null){//if existsif (ht[curr2.Next] != null){return true;}curr2 = curr2.Next;}return false;}算法2：把一个链表A接在另一个链表B的末尾，如果有环，则必然相交。如何判断有环呢?从A开始遍历，如果能回到A的表头，则肯定有环。 注意，在返回结果之前，要把刚才连接上的两个链表断开，恢复原状。
static bool JudgeIntersectLink2(Link head1, Link head2){bool exists = false;Link curr1 = head1;Link curr2 = head2;//goto the end of the link1while (curr1.Next != null){curr1 = curr1.Next;}//join these two linkscurr1.Next = head2;//iterate link2while (curr2.Next != null){if (curr2.Next == head2){exists = true;break;}curr2 = curr2.Next;}//recover original status, whether exists or notcurr1.Next = null;return exists;}算法3：如果两个链表的末尾元素相同，则必相交。
static bool JudgeIntersectLink3(Link head1, Link head2){Link curr1 = head1;Link curr2 = head2;//goto the end of the link1while (curr1.Next != null){curr1 = curr1.Next;}//goto the end of the link2while (curr2.Next != null){curr2 = curr2.Next;}if (curr1 != curr2)return false;elsereturn true;}新闻热点
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