java金字塔与多个2不工作

Question

我想创建一个金字塔并将每个数字乘以2，直到达到中间然后除以2，如下例所示。

然而，在编写我的代码后，我无法将数字加倍（i * 2），然后一旦到达中心，它就会除以2，直到它变为1

我的输出：

 package question5;

 public class Pyramid {

 public static void main(String[] args) {
    int x = 5;
    int rowCount = 1;

    System.out.println("Here Is Your Pyramid");

    //Implementing the logic

    for (int i = x; i > 0; i--)
    {
        //Printing i*2 spaces at the beginning of each row

        for (int j = 1; j <= i*2; j++)
        {
            System.out.print(" ");
        }

        //Printing j value where j value will be from 1 to rowCount

        for (int j = 1; j <= rowCount; j++)             
        {
        System.out.print(j+" ");
        }

        //Printing j value where j value will be from rowCount-1 to 1

        for (int j = rowCount-1; j >= 1; j--)
        {                    
        System.out.print(j+" ");             
        }                          

        System.out.println();

        //Incrementing the rowCount

        rowCount++;
    }
}
}

this is the weird output of it

Answer 1

这是有效的...您可以使用math.pow方法。

public class test {

     public static void main(String[] args) {
        int x = 5;
        int rowCount = 1;

        System.out.println("Here Is Your Pyramid");
        //Implementing the logic

        for (int i = x; i > 0; i--)
        {
            //Printing i*2 spaces at the beginning of each row
            for (int j = 1; j <= i*2; j++)
            {
                System.out.print(" ");
            }

            //Printing j value where j value will be from 1 to rowCount

            for (int j = 0; j <= rowCount-1; j++)             
            {
            System.out.printf("%2d", (int)Math.pow(2, j));  
            }

            //Printing j value where j value will be from rowCount-1 to 1

            for (int j = rowCount-1; j >= 1; j--)
            {                    
            System.out.printf("%2d", (int)Math.pow(2, j-1));
            }                          

            System.out.println();

            //Incrementing the rowCount

            rowCount++;

        }
    }
    }

Answer 2

现在您正在输出Math.floor(Math.random()*3) - 1 ，一个从j到1的号码。您想要输出 2 ^j-1 ，这很容易：5。

您还需要右键调整数字。然后您的打印声明变为：

1 << (j - 1)