Question

这是问题：给定两个数字H和N，请使用*绘制高度为H的N颗钻石。此外，对于N颗钻石中的每一个，也请使用o绘制另一颗高度为H-2的较小钻石，它们共享与先前的钻石位于同一中心。每个外部菱形都有三个连续的英文字母，按升序排列，以A开头，如下所示。

但是我只有字母，结果应该像

      *           *
    *   *       *   *
  *   O   *ABC *  O  *DEF
    *   *       *   *
      *           *

我不知道如何在*后添加字母而不弄乱菱形。

   import java.util.Scanner;
    public class manydiamond {
        public static void main(String[] args) {
            Scanner scn=new Scanner(System.in);
            System.out.println("Enter the height:");
            int h=scn.nextInt();
            System.out.print("How many diamonds?:");
            int a=scn.nextInt();
            char[] letter={'A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z'};
            //top of diamond
            for(int n=1;n<=h;n++) {         //which row
            for(int j=1;j<=a;j++)               //how many diamond
                {
                for(int k=1;k<=h-n;k++)         //left space
                {
                    System.out.print(" ");
                }
                for(int m=1;m<=2*n-1;m++)       //star
                {
                    if((m==1||m==2*n-1))
                    System.out.print("*");  
                    else if((n>=3)&&(m==3)||(m==2*n-3))
                    System.out.print("O");
                    else
                    System.out.print(" ");
                }
                for(int space1=1;space1<=-n+h;space1++)     //right space
                {   
                    System.out.print(" ");
                }   
                if(n==h)
                System.out.print(letter[0]);
            }
            System.out.println("");
            }
    int lh=h-1;                             //bottom row of diamond
            for(int i=1;i<=lh;i++)  {                       //which row
                for(int j=1;j<=a;j++) {                     //how many diamonds
                    for(int space=1;space<=i;space++)       //left space
                    {
                    System.out.print(" ");
                    }
                    for(int star=-2*i+(2*h-1);star>0;star--)        //star
                    {
                        if((star==1)||(star==-2*i+(2*h-1)))
                        System.out.print("*");
                        else if((i>=1&&(star==3)||(star==-2*i+(2*h-1)-2)))
                        System.out.print("O");
                        else
                        System.out.print(" ");
                    }
                    for(int space2=1;space2<=i;space2++)    //right space
                        System.out.print(" ");
                }
                System.out.println();
                }
        }
        }

Answer 1

对当前代码进行最少的不言自明的更改（请参见下面带有// CHANGED的行）：

通常，您只需要在这里和那里打印几个额外的空格，而不必总是打印出字母数组的第一个元素，即A。

import java.util.Scanner;

class ManyDiamond {
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    System.out.print("Enter the height: ");
    int h = sc.nextInt();
    System.out.print("How many diamonds: ");
    int a = sc.nextInt();
    sc.close();
    int lCurr = 0; // CHANGED
    char[] letter = { 'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R', 'S',
        'T', 'U', 'V', 'W', 'X', 'Y', 'Z' };
    // top of diamond
    for (int n = 1; n <= h; n++) { // which row
      for (int j = 1; j <= a; j++) { // how many diamond
        for (int k = 1; k <= h - n; k++) { // left space
          System.out.print(" ");
        }
        for (int m = 1; m <= 2 * n - 1; m++) { // star
          if ((m == 1 || m == 2 * n - 1)) {
            System.out.print("*");
          } else if ((n >= 3) && (m == 3) || (m == 2 * n - 3)) {
            System.out.print("O");
          } else {
            System.out.print(" ");
          }
        }
        for (int space1 = 1; space1 <= -n + h; space1++) { // right space
          System.out.print(" ");
        }
        if (n == h && j != a) { // CHANGED
          System.out.print(String.valueOf(letter[lCurr++]) + String.valueOf(letter[lCurr++]) + String.valueOf(letter[lCurr++])); // CHANGED
        } else {
          System.out.print("   "); // CHANGED
        }
      }
      System.out.println("");
    }
    int lh = h - 1; // bottom row of diamond
    for (int i = 1; i <= lh; i++) { // which row
      for (int j = 1; j <= a; j++) { // how many diamonds
        for (int space = 1; space <= i; space++) { // left space
          System.out.print(" ");
        }
        for (int star = -2 * i + (2 * h - 1); star > 0; star--) { // star
          if ((star == 1) || (star == -2 * i + (2 * h - 1))) {
            System.out.print("*");
          } else if ((i >= 1 && (star == 3) || (star == -2 * i + (2 * h - 1) - 2))) {
            System.out.print("O");
          } else {
            System.out.print(" ");
          }
        }
        for (int space2 = 1; space2 <= i; space2++) { // right space
          System.out.print(" ");
        }
        System.out.print("   "); // CHANGED
      }
      System.out.println();
    }
  }
}

示例用法1：

Enter the height: 5
How many diamonds: 4
    *           *           *           *
   * *         * *         * *         * *
  * O *       * O *       * O *       * O *
 * O O *     * O O *     * O O *     * O O *
* O   O *ABC* O   O *DEF* O   O *GHI* O   O *
 * O O *     * O O *     * O O *     * O O *
  * O *       * O *       * O *       * O *
   * *         * *         * *         * *
    *           *           *           *

用法示例2：

Enter the height: 3
How many diamonds: 2
  *       *
 * *     * *
* O *ABC* O *
 * *     * *
  *       *

请注意，我已将其留给您查看，以了解如何使其像示例输出一样以两倍间距排列。

在java中打印diamonds + alphabet的图形

1 个答案: