遍历链表一遍取出中间的节点

步骤，设立两个指针，同事指向头结点，一个指针一次跳一个节点，一个指针一次跳两个节点。当跳两个节点的指针到表尾时，跳一个节点的指针刚好到终点

int middle() const {if(!m_head || !m_tail)return 0;

if(m_head == m_tail)return m_head->m_data;Node *mid = m_head;for (Node *node = m_head; node->m_next&&node->m_next->m_next; node = node->m_next->m_next) {mid = mid->m_next;}return mid->m_data;}