可能过早地离开递归？

Question

我当前的递归函数在某种程度上起作用,但是当它回到堆栈时会自行消失.

void Graph::findPath( Room * curRoom )
{
if( curRoom -> myNumber == 0 )
{
    cout << "Outside.\n";
    //Escape the recursion!
}
else
{
    curRoom -> visited = true;

    if( curRoom -> North -> visited == false )
    {   

        escapePath[ _index ] = "North";
        cout << "_index: " << _index << "\n";
        ++_index;

        findPath( curRoom -> North );

        cout << "_index: " << _index << "\n";
        escapePath[ _index ] = "";
        --_index;
    }

    if( curRoom -> East -> visited == false )
    {   
        escapePath[ _index ] = "East";
        cout << "_index: " << _index << "\n";
        ++_index;

        findPath( curRoom -> East );

        cout << "_index: " << _index << "\n";
        escapePath[ _index ] = "";
        --_index;
    }

    if( curRoom -> South -> visited == false )
    {   
        escapePath[ _index ] = "South";
        cout << "_index: " << _index << "\n";
        ++_index;

        findPath( curRoom -> South );

        cout << "_index: " << _index << "\n";
        escapePath[ _index ] = "";
        --_index;
    }

    if( curRoom -> West -> visited == false )
    {   
        escapePath[ _index ] = "West";
        cout << "_index: " << _index << "\n";
        ++_index;

        findPath( curRoom -> West );

        cout << "_index: " << _index << "\n";
        escapePath[ _index ] = "";
        --_index;
    }
}
}

为了节省一些阅读,我们的想法是基本情况是找到0.否则它会尝试四个不同的基本方向,这是一个不同的编号房间.每次进行移动时,它都会添加对外部数组所做的移动,每次重新启动它都会从堆栈中移除该步骤.

我的问题是它在找到0时存储了正确的路径,但在备份的路上将其删除.

有没有办法逃脱它,比如休息.

没有任何例外

Answer 1

有一种方法可以使用异常退出递归,但我不推荐它.相反,修改您的函数以返回一个bool,指示您是否找到0并修改您的逻辑以从函数返回而不更改路径(如果已找到0).这是这个想法的例证:

bool Graph::findPath( Room * curRoom )
{
    if( curRoom -> myNumber == 0 )
    {
        cout << "Outside.\n";
        //Escape the recursion!
        return true;
    }
    // ...
    if (findPath( curRoom -> North ))
        return true;
    // ...
    return false;
}