如何使用for循环简化此代码？

Question

对于我的学校项目，我们正在创建一个纸牌游戏，我有一些硬编码的代码块。 我希望有人能够在我附加的代码中看到一个模式，并能够使用for循环对其进行压缩。

    // set the pointers of the cards in the pyramid
    card [0].setPtrs (card [1], card [2]);
    card [1].setPtrs (card [3], card [4]);
    card [2].setPtrs (card [4], card [5]);
    card [3].setPtrs (card [6], card [7]);
    card [4].setPtrs (card [7], card [8]);
    card [5].setPtrs (card [8], card [9]);
    card [6].setPtrs (card [10], card [11]);
    card [7].setPtrs (card [11], card [12]);
    card [8].setPtrs (card [12], card [13]);
    card [9].setPtrs (card [13], card [14]);
    card [10].setPtrs (card [15], card [16]);
    card [11].setPtrs (card [16], card [17]);
    card [12].setPtrs (card [17], card [18]);
    card [13].setPtrs (card [18], card [19]);
    card [14].setPtrs (card [19], card [20]);
    card [15].setPtrs (card [21], card [22]);
    card [16].setPtrs (card [22], card [23]);
    card [17].setPtrs (card [23], card [24]);
    card [18].setPtrs (card [24], card [25]);
    card [19].setPtrs (card [25], card [26]);
    card [20].setPtrs (card [26], card [27]);

    // set card X coords
    cardCentreX [0] = 800;
    cardCentreX [1] = 800 - card [1].getWidth () / 2;
    cardCentreX [2] = 800 + card [1].getWidth () / 2;
    cardCentreX [3] = cardCentreX [1] - card [1].getWidth () / 2;
    cardCentreX [4] = cardCentreX [2] - card [1].getWidth () / 2;
    cardCentreX [5] = cardCentreX [2] + card [1].getWidth () / 2;
    cardCentreX [6] = cardCentreX [3] - card [1].getWidth () / 2;
    cardCentreX [7] = cardCentreX [4] - card [1].getWidth () / 2;
    cardCentreX [8] = cardCentreX [5] - card [1].getWidth () / 2;
    cardCentreX [9] = cardCentreX [5] + card [1].getWidth () / 2;
    cardCentreX [10] = cardCentreX [6] - card [1].getWidth () / 2;
    cardCentreX [11] = cardCentreX [7] - card [1].getWidth () / 2;
    cardCentreX [12] = cardCentreX [8] - card [1].getWidth () / 2;
    cardCentreX [13] = cardCentreX [9] - card [1].getWidth () / 2;
    cardCentreX [14] = cardCentreX [9] + card [1].getWidth () / 2;
    cardCentreX [15] = cardCentreX [10] - card [1].getWidth () / 2;
    cardCentreX [16] = cardCentreX [11] - card [1].getWidth () / 2;
    cardCentreX [17] = cardCentreX [12] - card [1].getWidth () / 2;
    cardCentreX [18] = cardCentreX [13] - card [1].getWidth () / 2;
    cardCentreX [19] = cardCentreX [14] - card [1].getWidth () / 2;
    cardCentreX [20] = cardCentreX [14] + card [1].getWidth () / 2;
    cardCentreX [21] = cardCentreX [15] - card [1].getWidth () / 2;
    cardCentreX [22] = cardCentreX [16] - card [1].getWidth () / 2;
    cardCentreX [23] = cardCentreX [17] - card [1].getWidth () / 2;
    cardCentreX [24] = cardCentreX [18] - card [1].getWidth () / 2;
    cardCentreX [25] = cardCentreX [19] - card [1].getWidth () / 2;
    cardCentreX [26] = cardCentreX [20] - card [1].getWidth () / 2;
    cardCentreX [27] = cardCentreX [20] + card [1].getWidth () / 2;

    for (int i = 0 ; i < 28 ; i++)
    {
        cardCentreY [i] = 50;
        if (i == 1 || i == 2)
        {
            cardCentreY [i] = cardCentreY [0] + card [1].getHeight () / 2;
        }
        else if (i >= 3 && i <= 5)
        {
            cardCentreY [i] = cardCentreY [1] + card [1].getHeight () / 2;
        }
        else if (i >= 6 && i <= 9)
        {
            cardCentreY [i] = cardCentreY [3] + card [1].getHeight () / 2;
        }
        else if (i >= 10 && i <= 14)
        {
            cardCentreY [i] = cardCentreY [6] + card [1].getHeight () / 2;
        }
        else if (i >= 15 && i <= 20)
        {
            cardCentreY [i] = cardCentreY [10] + card [1].getHeight () / 2;
        }
        else if (i >= 21 && i <= 27)
        {
            cardCentreY [i] = cardCentreY [15] + card [1].getHeight () / 2;
        }
    }

Answer 1

您没有提供最低限度的工作示例，这使得很难测试任何可能的解决方案以确保它们与您的代码相同。 我用JavaScript而不是Java制作了一个简单版本的原型，这使我可以做一些简单的可视化操作来检查循环是否正常工作。 我假设您将能够将JavaScript逻辑转换回Java，而不会遇到太多麻烦。

这是循环（在JavaScript中）：

var x_row_base = 250; // You used 800 here
var y = 0;
var cards = Array(28);

// Set up card pyramid
for (var i = 0, row = 0; i < cards.length; row++) {
  var x = x_row_base;
  for (var col = 0; col < row; i++, col++) {
    cards[i] = make_card(i, row);
    // set x and y position
    cards[i].x = x;
    cards[i].y = y;
    // create links to "child" cards in next row
    if (i < 21) {
      cards[i].left_child_index = i + row;
      cards[i].right_child_index = i + row + 1;
    }
    // display the card
    draw_card(cards[i]);
    x += CARD_X_SPACING + CARD_X_PADDING;
  }
  y += CARD_Y_SPACING;
  x_row_base -= (CARD_X_SPACING + CARD_X_PADDING) / 2;
}

您可以使用下面的代码片段（ 或在jsfiddle上 ）尝试一下。 当您将鼠标悬停在父卡上时，可通过更改子卡的颜色来显示父卡/子卡链接。 这是笔记本电脑上的JavaScript可视化效果的屏幕截图：

 function draw_card(card) { var $card = $("<div/>"); $card.text(card.label); $card.addClass('card'); $card.css({ 'left': card.x, 'top': card.y, 'z-index': card.z }); $card.appendTo($('#my_canvas')); card.rect = $card; // last row doesn't actually have children if (card.left_child_index) { $card.mouseover(function() { cards[card.left_child_index].rect.addClass("child"); cards[card.right_child_index].rect.addClass("child"); }); $card.mouseout(function() { cards[card.left_child_index].rect.removeClass("child"); cards[card.right_child_index].rect.removeClass("child"); }); } } function make_card(label, z) { return { label: label, z: z }; } var CARD_X_PADDING = 4; var CARD_X_SPACING = 50; var CARD_Y_SPACING = 45; var x_row_base = 250; // You used 800 here var y = 0; var cards = Array(28); // Set up card pyramid for (var i = 0, row = 0; i < cards.length; row++) { var x = x_row_base; for (var col = 0; col < row; i++, col++) { cards[i] = make_card(i, row); // set x and y position cards[i].x = x; cards[i].y = y; // create links to "child" cards in next row if (i < 21) { cards[i].left_child_index = i + row; cards[i].right_child_index = i + row + 1; } // display the card draw_card(cards[i]); x += CARD_X_SPACING + CARD_X_PADDING; } y += CARD_Y_SPACING; x_row_base -= (CARD_X_SPACING + CARD_X_PADDING) / 2; } 

 .card { border: 1px solid black; text-align: center; position: absolute; width: 50px; padding: 20px 0px; background-color: gray; font-size: 15px; cursor: pointer; } .child { background-color: yellow; } 

 <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> <div id="my_canvas"></div> 

Answer 2

这是在第一个循环中复制模式的代码：

    public static void main(String[] args) {

    int offset = 0;
    int index = 0;
    for (int i=1; i<7; i++) {
        offset += i;
        for (int j=1; j<=i; j++) {
            card[index++].setPtrs(card[offset+j-1], card[offset+j]);
            System.out.printf("%d %d %d\n", index++, offset+j-1, offset+j);
        }
    }

}

结果：

0 1 2
1 3 4
2 4 5
3 6 7
4 7 8
5 8 9
6 10 11
7 11 12
8 12 13
9 13 14
10 15 16
11 16 17
12 17 18
13 18 19
14 19 20
15 21 22
16 22 23
17 23 24
18 24 25
19 25 26
20 26 27

Answer 3

我不会给你答案，因为自己动手做会使您成为一个更好的程序员，但这对您来说是个起点。

int i = 0;
int j = 5;
while (i < 21)
{
  for (int k = 0; k < j; k++)
  {
    card [i].setPtrs (card [k], card [k + 1]);
  }      
  j++;
  i++;
}

Answer 4

 // x - co ords
 cardCentreX [0] = 800;
 cardCentreX [1] = 800 - card [1].getWidth () / 2;
 cardCentreX [2] = 800 + card [1].getWidth () / 2;

 int last_count = 1;
 int curr_count = 0;
 int last_card = 0;
 boolean special = false;

 for (int i=3; i<= 27; i++){
     if (special){
         cardCentreX [i] = cardCentreX [last_card] + card [1].getWidth () / 2;
         special = false;
     } else {
          last_card++;
          cardCentreX [i] = cardCentreX [last_card] - card [1].getWidth () / 2;
          if (curr_count == last_count + 1){
              last_count = curr_count;
              curr_count = 0;
              special = true;
           }
     }

那应该工作。 对于x坐标。 这个想法是每x + 1个方程式都有一个重复的特殊属性吗？ 其中x是您到达特殊行之前经过的最后一行行数。

Answer 5

呈现的此for循环基于您已设定的数字模式，如果不清楚，请使用打印语句声明变量以了解工作原理（在注释中提供）：

 static int j = 1;
for(int 1 = 0; i <21; i++){

card [i].setPtrs (card [j], card [j+1]);
  if((j+1)%4 == 0) j-1; 

  j+2;
  counter++;
// System.out.println("i = " +i +", j = " +j +", j+1=" +(j+1));
    }

    // set card X coords
static int k = 1, counter = 1; // counter is a variable used for reference

cardCentreX [0] = 800;
cardCentreX [1] = 800 - card [1].getWidth () / 2;
cardCentreX [2] = 800 + card [1].getWidth () / 2;


for(int i=3; i < 28; i++){
   cardCentreX [i] = cardCentreX [k] - card [1].getWidth () / 2;
   if((i - k) == (counter+1)){ 
      k++; // increment card centre index
      counter++; 
   }
 // System.out.println("i = " +i, +", k = " +k, + ", counter = " +counter);
}

看看我如何缩短您的代码，您应该可以自己评估和缩短最后一个for循环。

Answer 6

cardCentrecount [0] = 800;
cardCentrecount [1] = 800 - card [1].getWidth () / 2;
cardCentrecount [2] = 800 + card [1].getWidth () / 2;

    int count = 3;
    int term = 2;
    for ( int i=3, j=1; i<=27; i++,j++ ) {

    if ( (j-1) == term ) {
        cardCentrecount[i] = cardCentrecount[ --j ] + card[1].getWidth() / 2;
        term += count;
        count ++;
    }
    else
        cardCentrecount[i] = cardCentrecount[j] + card[1].getWidth() / 2;
}