递归图

我理解递归的基础知识，但是当我遇到来自hackerrank的this one这样的问题时。我很快就对如何处理它感到困惑。递归图

基本上你必须绘制一个分形的ascii树（由字母Y组成），每层向下减半的Y数。我似乎无法让我的周围基本步头，我可以把它推广到其他层，如下图所示：

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如果任何人都可以给我正确的方向一推，这将不胜感激。

来源

2013-11-25 Bob Templ

您的图片与您的描述不符。您可以重新解决问题，因为每个后续行都有Y个字符数的一半。您的第一行将有32个Y字符，用空格分隔。您的下一行将有16个Y字符，位于第一行的Y字符之间。你可以找出其余的。 –

Ys翻倍，但我重新提出了这个问题，因为无论如何它会向下生成。 –

我想你需要生成确切的图像，你能说图像的大小：lines x cols？ – Pandrei

图像被以下列方式构造：

Y01 
    Y11 
--------- 
    Y02 
    Y12 
--------- 
    Y04 
    Y14 
--------- 
    Y08 
    Y18 
--------- 
    Y016 
    Y116

Y0x & Y1X用于字母Y：Y0x是枝条和Y1X是将trunc; x是形成字母所需的线的数量

例如， Y02 & Y12 - 2线，形成树枝和2号线，形成干线

你可以注意到，还有一个“规则”：当你开始绘制新字母（例如，你从YX2去Yx3）你复制最后一行。

编辑：让我们假设你有保存在一个矩阵char P[63][100] 我会写2功能图纸（不是最佳方法，但你可以优化它） - drawY - 平根据输入在Y字母 - drawPicutre - 画出“图层”Y

drawY(int startPos, int stopPos, int*line, int lvl, int count) 
{ 
    if(lvl>=count) 
    { 
    P[*line][(startPos+(stopPos-startPos))/2] = '1'; 
    P[*line + lvl][(startPos+(stopPos-startPos))/2 +count] = '1'; 
    P[*line + lvl][(startPos+(stopPos-startPos))/2 -count] = '1'; 
    (*line)--; 
    drawY(startPos,stopPos,line,lvl,count+1); 
    } 

} 

int line=63; 
drawY(0,100, &line, 16, 1); // draws the first, and largest Y

你只需要实现第二个函数drawPicture（...），它调用drawY。如果你愿意，你可以修改drawY。

来源

2013-11-25 21:34:54 Pandrei

个人我会从底部开始工作我的方式;但这只是一个偏好问题。 – Pandrei

我想，即使知道图像的结构，我很难将其分解为递归步骤和基本情况。 –

我有点得到你在说什么，除了我无法通过挑战的优点来声明变量，所以矩阵不在表格中。我很好地画出了整个结构 –

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