java - 寻找暂定大小的 ASCII 艺术程序的空白算法

Question

所以我们必须设计一本 ASCII 艺术书，我快完成了，但我想不出一件小事:“Building Java Programs”两边的空格

Here is what the book needs to look like

到目前为止，这是我的代码(为了便于帮助，我只展示了需要间距帮助的方法。假设 drawLine() 将虚线均匀地绘制到 SIZE 常数)

//constant SIZE = 8
public static void drawBottom() {
    //Dash line on top of the bottom portion of the book
    drawLine();
    //Printing first set of rightmost "/"'s
    for (int i = 1; i <= SIZE; i++)
        System.out.print("/");
    System.out.println();
    for (int i = 1; i <= SIZE / 2; i++) {
        //Leftmost pipe
        System.out.print("|");
    //    TO DO: Code label of book
    //    for (int j = 1; j <= ; j++) {
    //
    //    }
        //This loop is only here for example.
        //To show I can fill the space but need
        //the words in the space
        for (int j = 1; j <= SIZE * 3; j++) {
            System.out.print(" ");
        }
        //Rightmost pipe
        System.out.print("|");
        //"Pages" to right of label
        for (int j = 1; j <= -2 * i + (SIZE + 2); j++) {
            System.out.print("/");
        }
        //Move to draw next row
        System.out.println();
    }
    //Dash line on very bottom of entire drawing
    drawLine();
}

Here is my output (when SIZE = 8)

如何计算“Building Java Programs”文本 block 左右的间距？

我只知道当SIZE = 8时，两边各留一个空格

当SIZE = 10时，两边各有4个空格

当SIZE = 13时，两边各有8个空格

什么算法可以帮到我？

Answer 1

每一行可分为三个区域:第一个区域由位于索引A 到索引B 之间的空格组成。第二个区域包含从包含索引C 到包含索引D 的文本。第三个区域再次由位于从包含索引 E 到包含索引 F 的空间组成。侧翼管道位于索引 0 和 width + 1 处:

|A......BC......DE......F|

第一区和第三区的长度为width/2 - text/2，其中text表示文本的长度。

那么索引是:

Index A: 1
Index B: width/2 - text/2  
Index C: B + 1
Index D: width/2 + text/2
Index E: D + 1
Index F: width

在循环中，可以在相应的区域显示所需的字符:

// Code: label of book
int width = 3 * SIZE; 
width = (width % 2 == 0) ? width : width - 1;           // if the width is odd, choose the next smallest even number 
String text = "Building Java Programs";
for(int j = 1; j <= width; j++) {
    if (j <= width / 2 - text.length() / 2) {           // j <= Index B
        System.out.print(" ");
    }
    else if (j >= width / 2 + text.length() / 2 + 1) {  // j >= Index E = Index D + 1    
        System.out.print(" ");
    }
    else {
        System.out.print(text);
        j = width / 2 + text.length() / 2;              // j = Index D
    }
}

当然，第一个和第二个if语句也可以结合起来实现。

输出(SIZE = 8，width = 24):

Without text...
|                        |////////
|                        |//////
|                        |////
|                        |//

With text...
| Building Java Programs |////////
| Building Java Programs |//////
| Building Java Programs |////
| Building Java Programs |//

两边各留一个空格 (1 + 22 + 1 = 24)。

输出(SIZE = 10，width = 30):

Without text...
|                              |//////////
|                              |////////
|                              |//////
|                              |////
|                              |//

With text...
|    Building Java Programs    |//////////
|    Building Java Programs    |////////
|    Building Java Programs    |//////
|    Building Java Programs    |////
|    Building Java Programs    |//

两边各有四个空格 (4 + 22 + 4 = 30)。

输出(SIZE = 13，width = 38):

Without text...
|                                      |/////////////
|                                      |///////////
|                                      |/////////
|                                      |///////
|                                      |/////
|                                      |///

With text...
|        Building Java Programs        |/////////////
|        Building Java Programs        |///////////
|        Building Java Programs        |/////////
|        Building Java Programs        |///////
|        Building Java Programs        |/////
|        Building Java Programs        |///

两边各有八个空格 (8 + 22 + 8 = 38)。