带数字的空钻石形状

Question

所以有人问我这个问题，我只能解决代码的顶部，我卡在了底部。

Write a Java program called EmptyDiamond.java that contains a method that takes an integer n and prints a empty rhombus on 2n − 1 lines as shown below. Sample output where n = 3:
  1
 2 2
3   3
 2 2
  1

到目前为止，这是我的代码：

public static void shape(int n) {
    //TOP PART
    for (int i = 1; i <= (n - 1); i++) {
        System.out.print(" ");
    }
    System.out.println(1);
    for (int i = 2; i <= n; i++) {
        for (int j = 1; j <= (n - i); j++) {
            System.out.print(" ");
        }
        System.out.print(i);
        for (int j = 1; j <= 2 * i - n + 1; j++) {
            System.out.print(" ");
        }
        System.out.println(i);
    }

    //BOTTOM PART (The messed up part)
    for (int i = n + 1; i <= 2 * n - 2; i++) {
        for (int j = 1; j <= n - i; j++) {
            System.out.print(" ");
        }
        System.out.print(i);
        for (int j = 1; j <= n; j++) {
            System.out.print(" ");
        }
        System.out.print(i);
    }
    for (int i = 1; i <= (n - 1); i++) {
        System.out.print(" ");
    }
    System.out.println(1);
}

public static void main(String[] args) {
    shape(4);
}

Answer 1

打印空钻石的程序如下：

int n = 3; //change the value of n to increase the size of diamond
int upperCount = 1;
for (int i = n; i >= 1; i--) {
    for (int j = i; j >= 1; j--) {
        System.out.print(" ");
    }
    System.out.print(upperCount);
    for (int j = 0; j <= upperCount - 2; j++) {
        System.out.print(" ");
    }
    for (int j = 0; j <= upperCount - 2; j++) {
        System.out.print(" ");
    }
    if (upperCount != 1) {
        System.out.print(upperCount);
    }
    upperCount++;
    System.out.print("\n");
}

int lowerCount = n - 1;
for (int i = 1; i <= n - 1; i++) {
    for (int j = 0; j <= i; j++) {
        System.out.print(" ");
    }
    System.out.print(lowerCount);
    for (int j = 0; j <= lowerCount - 2; j++) {
        System.out.print(" ");
    }
    for (int j = 0; j <= lowerCount - 2; j++) {
        System.out.print(" ");
    }
    if (lowerCount != 1) {
        System.out.print(lowerCount);
    }
    lowerCount--;
    System.out.print("\n");
}

在代码的底部进行以下更改：

int lowerCount = n - 1;
for (int i = n - 1; i >= 2; i--) {
    for (int j = 1; j <= (n - i); j++) {
        System.out.print(" ");
    }
    System.out.print(i);

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

Answer 2

您可以使用两个嵌套的 for 循环在从 -n 到 n 的行和列上打印 带有数字 的空菱形。当iAbs + jAbs == n:

时得到菱形

int n = 2;
for (int i = -n; i <= n; i++) {
    // absolute value of 'i'
    int iAbs = Math.abs(i);
    for (int j = -n; j <= n; j++) {
        // absolute value of 'j'
        int jAbs = Math.abs(j);
        // empty diamond shape
        System.out.print(iAbs + jAbs == n ? jAbs + 1 : " ");
        if (j < n) {
            System.out.print(" ");
        } else {
            System.out.println();
        }
    }
}

输出：

可以分别定义width和height:

int m = 4;
int n = 2;
int max = Math.max(m, n);
for (int i = -m; i <= m; i++) {
    // absolute value of 'i'
    int iAbs = Math.abs(i);
    for (int j = -n; j <= n; j++) {
        // absolute value of 'j'
        int jAbs = Math.abs(j);
        // empty diamond shape
        System.out.print(iAbs + jAbs == max ? jAbs + 1 : " ");
        if (j < n) {
            System.out.print(" ");
        } else {
            System.out.println();
        }
    }
}

输出：

^{另请参阅：

•

• How to print a diamond of random numbers?}

Answer 3

java-11

使用作为 Java-11 的一部分引入的 String#repeat，您可以使用 单循环 .

public class Main {
    public static void main(String[] args) {
        int n = 3;
        for (int i = 1 - n; i < n; i++) {
            int x = Math.abs(i);
            System.out.println(" ".repeat(x) + (n - x)
                    + " ".repeat(Math.abs((n - x) * 2 - 3))
                    + ((i == 1 - n || i == n - 1) ? "" : (n - x)));
        }
    }
}

输出：

您只需将 space 的数量增加一个字符即可打印钻石的变体：

public class Main {
    public static void main(String[] args) {
        int n = 3;
        for (int i = 1 - n; i < n; i++) {
            int x = Math.abs(i);
            System.out.println("  ".repeat(x) + (n - x)
                    + "  ".repeat(Math.abs((n - x) * 2 - 3))
                    + ((i == 1 - n || i == n - 1) ? "" : (n - x)));
        }
    }
}

输出：

Answer 4

备选方案：

public static void main(String[] args) {
  int n = 7;
  for (int i = -n; i <= n; i++) {
    for (int j = -n; j <= n; j++) {
      // edge of the diamond
      int edge = Math.abs(i) + Math.abs(j);
      // diamond shape with numbers
      if (edge == n) System.out.print(n - Math.abs(i) + 1);
      // beyond the edge && in chessboard order || vertical borders
      else if (edge > n && (i + j) % 2 != 0 || Math.abs(j) == n)
        System.out.print("*");
      // empty part
      else System.out.print(" ");
    }
    System.out.println();
  }
}

输出：

** * * 1 * * **
* * * 2 2 * * *
** * 3   3 * **
* * 4     4 * *
** 5       5 **
* 6         6 *
*7           7*
8             8
*7           7*
* 6         6 *
** 5       5 **
* * 4     4 * *
** * 3   3 * **
* * * 2 2 * * *
** * * 1 * * **

^{另请参阅：How to print a given diamond pattern in Java?}

Answer 5

可能有点晚了，但是因为消息的底部只是镜像的第一部分，所以您可以使用 Stack 以相反的顺序打印消息：

public static void main(String[] args) {
    int maxNumber = 3;
    Stack<String> message = new Stack<>();
    // upper part
    for (int row = 0; row < maxNumber; row++) {
        int prefix = maxNumber - (row + 1);
        int spaces = row >= 2 ? row * 2 - 1 : row;

        String line = getLine(row, prefix, spaces);
        System.out.println(line);
        if (row != maxNumber - 1)
            message.add(line);
    }
    // bottom part
    while (!message.isEmpty())
        System.out.println(message.pop());
}

public static String getLine(int row, int prefix, int spaces) {
    StringBuilder line = new StringBuilder("_".repeat(prefix));
    line.append(row + 1);
    if (row != 0) {
        line.append("_".repeat(spaces));
        line.append(row + 1);
    }
    return line.toString();
}

输出：

__1
_2_2
3___3
_2_2
__1

你当然可以使用任何你想要填充堆栈的方法（即生成消息的上半部分），就像这个问题建议的方法一样。我描述的这个上半部分包含第一行（含）到中间行（不含）。

Answer 6

解决方案：Java 名为 EmptyDiamond.java 的程序包含一个采用整数 n 并在 2n − 1 行打印空菱形的方法。

public class EmptyDiamond {
    public static void main(String[] args) {
        shape(3); // Change n to increase size of diamond
    }

    public static void shape(int n) {
        int max = 2 * n - 1; // length of the diamond - top to bottom
        int loop = 0; // with of each loop. initialized with 0
        for (int i = 1; i <= max; i++) {
            int val = 0;
            if (i <= n) {
                loop = n + i - 1;// e.g. when i = 2 and n = 3 loop 4 times
                val = i; // value to be printed in each loop ascending
            } else {
                loop = n + (max - i); //e.g. when i = 4 and n = 3 loop 4 times
                val = max - i + 1; // value to be printed in each loop descending
            }
            for (int j = 1; j <= loop; j++) {
                // (value end of loop)
                // || (value in the beginning  when i <= n)
                // || (value in the beginning  when i > n)
                if (j == loop
                        || j == (n - i + 1)
                        || j == (n - val + 1)) {
                    System.out.print(val); // Print values
                } else {
                    System.out.print(" "); // Print space
                }
            }
            System.out.println(); // Print next line
        }
    }
}

当n = 3时输出：

Answer 7

仅流函数：

public static void printEmptyDiamond(int n) {
   IntStream.range(1, 2*n)
        .map(i-> i > n ? 2*n-i : i)
        .mapToObj(i -> " ".repeat(n-i) + i + (i>1 ? " ".repeat(2*(i-1)-1)+i : ""))
        .forEach(System.out::println);
}

示例输出 (printEmptyDiamond(7)):

      1
     2 2
    3   3
   4     4
  5       5
 6         6
7           7
 6         6
  5       5
   4     4
    3   3
     2 2
      1

有说明：

public static void printEmptyDiamond(int n) {
   IntStream.range(1, 2*n)
        .map(i-> i > n? 2*n-i : i) // numbers from 1 to n ascending, then descending to 1 again
        .mapToObj(i -> " ".repeat(n-i) // leading spaces
            + i // leading number
            + (i>1 ? // only when number is > 1
                " ".repeat(2*(i-1)-1) // middle spaces
                 + i // trailing number
            : ""))
        .forEach (System.out::println);
}

Answer 8

我这样做是为了好玩，这是代码：

import java.util.Scanner;

public class Diamond {
    public static void main(String[] args) {
        Scanner read = new Scanner(System.in);
        int num = read.nextInt();
        read.nextLine();
        //TOP
        for(int i = 1;i<=num;i++) {
            //LEFT
            for(int k = i; k<num;k++) {
                if ( k % 2 == 0 ) {
                    System.out.print(" ");
                }
            else {
                System.out.print(" ");
            }
        }
        if(i>1) {
            for(int j =1;j<=i;j++) {
                if (j==1 || j== i) {
                    for(int u=0;u<j;u++) {
                        System.out.print(" ");
                    }
                    System.out.print(i);
                    
                }
                else {
                    System.out.print(" ");
                }
            }
            System.out.println("");
        }
        else {
            System.out.println("  "+i);
        }
    }
    //BOTTOM
    for(int i = num-1;i>0;i--) {
        for(int k = i; k<num;k++) {
            if ( k % 2 == 0 ) {
                System.out.print(" ");
            }
            else {
                System.out.print(" ");
            }
        }
        if(i>1) {
            for(int j =1;j<=i;j++) {
                if (j==1 || j== i) {
                    for(int u=0;u<j;u++) {
                        System.out.print(" ");
                    }
                    System.out.print(i);
                }
                else {
                    System.out.print(" ");
                }
            }
            System.out.println("");
        }
        else {
            System.out.println(" "+i);
        }
    }
}
}

输出：

    7
        1
      2  2
     3    3
    4      4
   5        5
  6          6
 7            7
  6          6
   5        5
    4      4
     3    3
      2  2
       1

看过其他答案后，我可以跳过一大堆循环。只是决定在最后即兴发挥并尽快完成。

带数字的空钻石形状

Empty diamond shape with numbers

java

ascii-art

java-11