题目：在本题中， 单链表可能有环， 也可能无环。 给定两个单链表的头节点head1和head2， 这两个链表可能相交， 也可能不相交。 请实现一个函数，如果两个链表相交， 请返回相交的第一个节点； 如果不相交， 返回null即可。
思路:
1.先判断链表是否有环
2.然后再根据下面的情况，分别讨论。

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public static class Node {
        public int value;
        public Node next;

        public Node(int data) {
            this.value = data;
        }
    }

//主要功能函数，判断二者有没有连接点
public static Node getIntersect Node(Node head1, Node head2) {
        if (head1 == null || head2 == null) {
            return null;
        }
        Node loop1 = getLoopNode(head1);
        Node loop2 = getLoopNode(head2);
        if (loop1 == null && loop2 == null) {
            return noLoop(head1, head2);
        }
        if (loop1 != null && loop2 != null) {
            return bothLoop(head1, loop1, head2, loop2);
        }
        return null;
    }


//判断一个链表中是否有环
public static Node getLoopNode(Node head) {
        if (head == null || head.next == null || head.next.next == null) {
            return null;
        }
        Node n1 = head.next; // n1 -> slow
        Node n2 = head.next.next; // n2 -> fast
        while (n1 != n2) {
            if (n2.next == null || n2.next.next == null) {
                return null;
            }
            n2 = n2.next.next;
            n1 = n1.next;
        }
        n2 = head; // n2 -> walk again from head
        while (n1 != n2) {
            n1 = n1.next;
            n2 = n2.next;
        }
        return n1;
    }

//如二者都没有环，判断他们是否有相交节点
public static Node noLoop(Node head1, Node head2) {
        if (head1 == null || head2 == null) {
            return null;
        }
        Node cur1 = head1;
        Node cur2 = head2;
        int n = 0;
        while (cur1.next != null) {
            n++;
            cur1 = cur1.next;
        }
        while (cur2.next != null) {
            n--;
            cur2 = cur2.next;
        }
        if (cur1 != cur2) {
            return null;
        }
        cur1 = n > 0 ? head1 : head2;
        cur2 = cur1 == head1 ? head2 : head1;
        n = Math.abs(n);
        while (n != 0) {
            n--;
            cur1 = cur1.next;
        }
        while (cur1 != cur2) {
            cur1 = cur1.next;
            cur2 = cur2.next;
        }
        return cur1;
    }

//若二者都有环，判断他们是否相交

public static Node bothLoop(Node head1, Node loop1, Node head2, Node loop2) {
        Node cur1 = null;
        Node cur2 = null;
        if (loop1 == loop2) {
            cur1 = head1;
            cur2 = head2;
            int n = 0;
            while (cur1 != loop1) {
                n++;
                cur1 = cur1.next;
            }
            while (cur2 != loop2) {
                n--;
                cur2 = cur2.next;
            }
            cur1 = n > 0 ? head1 : head2;
            cur2 = cur1 == head1 ? head2 : head1;
            n = Math.abs(n);
            while (n != 0) {
                n--;
                cur1 = cur1.next;
            }
            while (cur1 != cur2) {
                cur1 = cur1.next;
                cur2 = cur2.next;
            }
            return cur1;
        } else {
            cur1 = loop1.next;
            while (cur1 != loop1) { 
                if (cur1 == loop2) {
                    return loop1;
                }
                cur1 = cur1.next;
            }
            return null;
        }
    }