如何判断连个单链表（无环）是否交叉

class LNode:
    def __init__(self):
        self.data = None
        self.next = None


def IsIntersect(head1, head2):
    """
    方法功能：判断两个链表是否相交，如果相交找出交点
    :param head1: 第一个链表
    :param head2: 第二个链表
    :return: 如果不相交返回None, 如果相交返回相交结点
    """

    if head1 is None or head1.next is None or head2 is None or head2.next is None or head1 == head2:
        return None
    temp1 = head1.next
    temp2 = head2.next
    n1, n2 = 0, 0
    # 遍历head1，找到尾结点，同时记录head1的长度
    while temp1.next is not None:
        temp1 = temp1.next
        n1 += 1
    # 遍历head2，找到尾结点，同时记录head2的长度
    while temp2.next is not None:
        temp2 = temp2.next
        n2 += 1
    if temp1 == temp2:
        # 长链表线走|n1-n2|步
        if n1 > n2:
            while n1 - n2 > 0:
                head1 = head1.next
                n1 -= 1
        if n2 > n1:
            while n2 - n1 >0:
                head2 = head2.next
                n2 -= 1
        # 两个链表同时前进，找出相同的结点
        while head1 != head2:
            head1 = head1.next
            head2 = head2.next
        return head1
    # head1 与head2 是没有相同的尾结点
    else:
        return None


if __name__ == '__main__':
    i = 1
    # 链表头结点
    head1 = LNode()
    head2 = LNode()
    tmp = None
    cur = head1
    p = None
    # 构造第一个链表
    while i < 8:
        tmp = LNode()
        tmp.data = i
        cur.next = tmp
        cur = tmp
        if i == 5:
            p = tmp
        i += 1

    cur = head2
    # 构造第二个链表
    i = 1
    while i < 5:
        tmp = LNode()
        tmp.data = i
        cur.next = tmp
        cur = tmp
        i += 1
    # 是它们相交于点5
    cur.next = p
    interNode = IsIntersect(head1, head2)
    if interNode is None:
        print('这两个链表不相交')
    else:
        print('这两个链表相交点：' + str(interNode.data))

运行结果如下：
这两个链表相交点：5