Question

我目前正在尝试制作一个形状有点麻烦我的代码是：

winsound.PlaySound(G5.wav)

一切正常，除了我的上一个方法“fancySquare”没有打印出我想要的方式。如果用户输入为9：

，它当前正在打印此形状

winsound

它应该是这样的

import java.util.Scanner;
public class Problem2
{
   public static void main(String[] args)
   {
      Scanner input = new Scanner(System.in);

      int n = 0;

      do
      {
         System.out.print("Enter an odd integer greater than or equal to 5: ");
         n = input.nextInt();

      }while(n < 5 || n % 2 == 0);

      boxWithMinorDiagonal(n);

      rightTriangle(n);

      printNLetter(n);

      fancySquare(n);

   }


   public static void fancySquare(int n)
   {
      System.out.println();

      int tempRow = 2;
      int tempCol = 2;
      int tempColDec = n - 2;
      int tempRowInc = 2;

      for(int row = 1; row <= n; row++)
      {
         for(int col = 1; col <= n; col++)
         {
            if(row == 1 || row == n || col == 1 || col == n)
            {
               if(row == 1 && col == 1 || row == 1 && col == n || row == n && col == 1 || row == n && col == n)
               {
                  System.out.print("@");
               }
               else
               {
                  System.out.print("*");
               }
            }

            else if(tempRow == row && tempCol == col)
            {
               System.out.print("+");
               tempCol++;
               tempRow++;
            }

            else
            {
               System.out.print(" ");
            }
            if(row == tempRowInc && col == tempColDec)
            {
               System.out.print("+");
               tempColDec--;
               tempRowInc++;
            }


         }
         System.out.println();
      }
   }
}

Answer 1

尝试使用命名布尔值进行检查，以使其更容易理解。更常见的是使用零基索引，但我坚持你的版本，使它更相似，更容易匹配。

public static void fancySquare(int n) {
    System.out.println();

    for (int row = 1; row <= n; row++) {
        for (int col = 1; col <= n; col++) {

            final boolean firstRow = row == 1;
            final boolean lastRow = row == n;
            final boolean firstCol = col == 1;
            final boolean lastCol = col == n;

            // calc the center
            final boolean center = row == (n + 1) / 2 && col == (n + 1) / 2;
            // calc the marker
            final boolean marker = row == col || n - row + 1 == col;

            // we need the @ marker at the corners (1,1) || (1,n) || (n,1) || (n,n)
            if ((firstRow || lastRow) && (firstCol || lastCol)) {
                System.out.print("@");
            // we need the * in the first & last column/row which is no corner
            } else if (firstRow || lastRow || firstCol || lastCol) {
                System.out.print("*");
            // we need a @ also in the center
            } else if (center) {
                System.out.print("@");
            // we need the plus on the diagonals
            } else if (marker) {
                System.out.print("+");
            // instead we need a space
            } else {
                System.out.print(" ");
            }
        }
        System.out.println();
    }
}

嵌套循环来制作形状

1 个答案: