在java中將兩個相互遞歸的方法轉換爲單一的遞歸方法？

Question

我需要採用兩個相互遞歸方法的程序，並修改程序，使其包含單個遞歸方法。根據我的理解，我需要將遞歸調用按照調用順序放置在單個方法中，以結合這兩種遞歸方法。問題是有4個整數通過方法傳遞，第一個方法調用第二個方法兩次，第二個方法調用第一個方法兩次。

這是原來的代碼：

public void drawHorizontal(Graphics graphics, double xMid, double yMid, double length) 
{ 
    // find left endpoint 
    double x1 = xMid - (length/2); 
    double y1 = yMid; 

    // find right endpoint 
    double x2 = xMid + (length/2); 
    double y2 = yMid; 

    if (length > 5) 
    { 
    // draw a line from (x1,y1) to (x2,y2) 
    graphics.drawLine((int) x1, (int) y1, (int) x2, (int) y2); 

    // draw a vertical line with left end of horizontal as midpoint of new line 
    drawVertical(graphics, x1, y1, (length)); 

    // draw a vertical line with right endof horizontal as midpoint of new line 
    drawVertical(graphics, x2, y2, (length)); 
    }  

} // end drawHorizontal() 

public void drawVertical(Graphics graphics, double xMid, double yMid, double length) 
{ 
    // find upper endpoint 
    double x1 = xMid; 
    double y1 = yMid - (length/2); 

    // right lower endpoint 
    double x2 = xMid; 
    double y2 = yMid + (length/2); 


    if (length > 5) 
    { 
    // draw a line from (x1,y1) to (x2,y2) 
    graphics.drawLine((int) x1, (int) y1, (int) x2, (int) y2); 

    // draw a 1/2 size horizontal line with top end of vertical as midpoint of new line 
    drawHorizontal(graphics, x1, y1, (length/2)); 

    // draw a 1/2 horizontal line with bottom end of vertical as midpoint of new line 
    drawHorizontal(graphics, x2, y2, (length/2)); 
    } 

} // end drawVertical() 

下面是我最近修改的代碼。它的醜陋我知道，但我不知道如何獨立調整x和y座標。我試圖通過創建更多變量來解決這個問題，但我不禁覺得我只是在做更多的工作。我能找到的最接近的堆棧問題是this。我從11號開始就已經在這裏了，現在是4:15。向正確的方向微調將不勝感激，謝謝你的時間。

編輯*感謝您的快速回復，我很感激。我知道，以這種方式破解相互遞歸的方法似乎違反直覺，但我是編程和Java的新手，所以我正在探索解決問題的不同方法。這是我最終把它分解，並且運行良好。感謝您的時間。

修改後的代碼：

public void Total(Graphics graphics, boolean type, double xMid, double yMid, double length) { 

    double x1; 
    double y1; 

    // find right endpoint 
    double x2; 
    double y2; 

    if (type == false) { 
     // find left endpoint 
     x1 = xMid - (length/2); 
     y1 = yMid; 

     // find right endpoint 
     x2 = xMid + (length/2); 
     y2 = yMid; 

     if (length > 5) { 
      // draw a line from (x1,y1) to (x2,y2) 
      graphics.drawLine((int) x1, (int) y1, (int) x2, (int) y2); 

      // draw a vertical line with left end of horizontal as midpoint of new line 
      Total(graphics, true, x1, y1, (length)); 

      // draw a vertical line with right endof horizontal as midpoint of new line 
      Total(graphics, true, x2, y2, (length)); 
     } 
    } else { 

     // find upper endpoint 
     x1 = xMid; 
     y1 = yMid - (length/2); 

     // right lower endpoint 
     x2 = xMid; 
     y2 = yMid + (length/2); 

     if (length > 5) { 
      // draw a line from (x1,y1) to (x2,y2) 
      graphics.drawLine((int) x1, (int) y1, (int) x2, (int) y2); 

      // draw a 1/2 size horizontal line with top end of vertical as midpoint of new line 
      Total(graphics, false, x1, y1, (length/2)); 

      // draw a 1/2 horizontal line with bottom end of vertical as midpoint of new line 
      Total(graphics, false, x2, y2, (length/2)); 
     } 
    } 
} 

Answer 1

您的要求是奇怪的，相互遞歸通常有助於閱讀...而行，有（至少）一個用於該一般方法。如果您有：

f(f_params) { 
    … 
    g(g_params…); 
    … 
} 
g(g_params) { 
    … 
    f(f_params…); 
    … 
} 

可能被轉化這樣的：

fg(what,f_params,g_params) { 
    if (what==FUNC_F) { 
    … 
    fg(FUNC_G,f_params,g_params); 
    … 
    } else if (what==FUNC_G) { 
    … 
    fg(FUNG_F,f_params,g_params); 
    … 
    } 
} 

這是一個有點複雜，如果你需要管理不同的返回類型：只需添加相應的返回值有些out參數/類型。