Question

对于我的学校项目，我们正在创建一个纸牌游戏，我有一些我硬编码的巨大代码块。我希望有人能够在我附加的代码中看到一个模式，并且能够用for循环来压缩它。

    // 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

您没有提供最小的工作示例，因此很难测试任何可能的解决方案，以确保它们与您的代码执行相同的操作。我用JavaScript而不是Java原型化了一个简单的版本，这让我可以做一些简单的可视化，作为我的循环正常工作的健全性检查。我假设您可以将JavaScript逻辑转换回Java而不会有太多麻烦。

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段（or on jsfiddle）进行试用。当您将鼠标悬停在父级上时，通过更改子卡的颜色来显示父/子卡链接。以下是我的笔记本电脑上JavaScript可视化的截图：

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

以下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：

Answer 3

我不会给你答案，因为自己解决这个问题会让你成为一个更好的程序员，但这里有一个起点。

sentences = ['The number of adults who read at least one novel within the past 12 months fell to 47%.', 'Fiction reading rose from 2002 to 2008.', 'The decline in fiction reading last year occurred mostly among men.', 'Women read more fiction.', '50% more.', 'Though it decreased 10% over the last decade.', 'Men are more likely to read nonfiction.', 'Young adults are more likely to read fiction.', 'Just 54% of Americans cracked open a book of  any kind last year.', 'But novels have suffered more than nonfiction.']

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;
           }
     }

那应该有用。对于x坐标。这个想法是每个x + 1方程都有一个反复出现的特殊属性？其中x是在到达特殊线之前经过的最后一行数。

Answer 5

此for循环基于您已设定的数字模式，如果不清楚则使用变量的print语句来理解工作（在评论中提供）：

 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);
}

看看我如何缩短你的代码，你应该能够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;
}

如何使用for循环简化此代码？

6 个答案: