1
2

输入: 1->2->3->4->5->NULL, m = 2, n = 4
输出: 1->4->3->2->5->NULL

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impl Solution {
    pub fn reverse_between(head: Option<Box<ListNode>>, m: i32, n: i32) -> Option<Box<ListNode>> {
        if head.is_none() || n <= m  {
            return head;
        }

        let mut dummy_list: Option<Box<ListNode>> = Some(Box::new(ListNode::new(0)));
        let mut tail_ptr = &mut dummy_list;
        let mut tmp_list: Option<Box<ListNode>> = Some(Box::new(ListNode::new(0)));
        let mut i = 0;
        let mut next = head;
        while let Some(mut cur_node) = next.take() {
            i += 1;
            next = cur_node.next.take(); //取下当前节点下一个结点，将所有权转交给head
            if i < m {
                tail_ptr.as_mut().unwrap().next = Some(cur_node);
                tail_ptr = &mut tail_ptr.as_mut().unwrap().next;
            } else if i >= m && i <= n {
                cur_node.next = tmp_list.as_mut().unwrap().next.take();
                tmp_list.as_mut().unwrap().next = Some(cur_node);
                //
                if i == n {
                    //contact rev_head.next to tail
                    let mut tmp_node = tmp_list.as_mut().unwrap().next.take();
                    while let Some(mut node) = tmp_node {
                        tmp_node = node.next.take();
                        tail_ptr.as_mut().unwrap().next = Some(node);
                        tail_ptr = &mut tail_ptr.as_mut().unwrap().next;
                    }
                    //如果剩余结点不为空
                    if next.is_some() {
                        tail_ptr.as_mut().unwrap().next = next;
                    }
                    break;
                }
            } else {
                break;
            }
        }

        dummy_list.unwrap().next
    }
}