架构师系列 9: 找出单向链表合并节点

def check(head_a, head_b):    if head_a is None or head_b is None or head_a.next is None or head_b.next is None:        return None    # 空间间复杂度O(n)    hashset = set()    hashset.clear()    p = head_a.next    # 时间复杂度O(n)    while p.next is not None:        hashset.add(p.next)        p = p.next    print("link_a addr: {0}".format(hashset))    p = head_b.next    while p is not None:        if p.next in hashset:            print("link_b addr 相交节点地址: {0} 值：{1}".format(p.next, p.data))            return p.data  # 相交元素        print("link_b addr: {0}".format(p.next))        p = p.next    print("不相交")    return None