Question

寻求自然数n时，我想以以下格式打印到控制台：

                1
              2 1
            3 2 1
          4 3 2 1
        5 4 3 2 1
          .
          .
          .
n . . . 5 4 3 2 1

输入4，这是我到目前为止的内容：

我想在数字之间添加一个空格。这是我的代码：

import java.util.Scanner;
public class PatternTwo {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int userInput;
        System.out.println("Please enter a number 1...9 : ");
        userInput = in.nextInt();
        String s="";
        int temp = userInput;
        for(int i=1; i<=userInput; i++ ) {

            for (int k= userInput; k>=i; k-- ) {
                System.out.printf(" ");
            }

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


            System.out.println("");
        }

    }

}

Answer 1

在要打印的数字前面添加一个空格，并将上面的空格加倍，以使其不是金字塔。像这样：

import java.util.Scanner;
public class PatternTwo {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int userInput;
        System.out.println("Please enter a number 1...9 : ");
        userInput = in.nextInt();
        String s="";
        int temp = userInput;
        for(int i=1; i<=userInput; i++ ) {

            for (int k= userInput; k>i; k-- ) { // <- corrected condition
                System.out.printf("  ");
            }

            for(int j = i; j>=1; j-- ) {
                System.out.print(j);

                // check if not 1 to avoid a trailing space
                if (j != 1) {
                    System.out.print(" ");
                }
            }


            System.out.println("");
        }

    }

}

编辑

感谢/u/shash678，我更正了解决方案，以删除所有不必要或错误的空格

Answer 2

一个更干净的解决方案如何避免使用嵌套的for循环呢？

public static void main(final String[] args) {
    final Scanner scanner = new Scanner(System.in);
    System.out.print("Please enter a number 1...9 : ");
    final int n = scanner.nextInt();

    final ArrayList<String> suffixes = new ArrayList<>();
    for (int i = 1; i <= n; i++) {
        suffixes.add(0, i + " ");

        final String prefix = String.join("", Collections.nCopies(n - i, "  "));
        final String suffix = String.join("", suffixes);

        System.out.println(prefix + suffix);
    }
}

Answer 3

以下是对答案的修改，当k等于i时，通过修改k个for循环退出条件，并且类似地当j等于{{时，不会打印不必要的多余空白1}}，将其分开处理。

主要的总体变化是，在k循环中，您需要打印2个空格而不是1个空格才能实现所需的右侧对齐：

用法示例：

import java.util.Scanner;

class PatternTwo {

  private static void printPatternTwo(int n) {
    for (int i = 1; i <= n; i++) {
      for (int k = n; k > i; k--) {
        System.out.print("  ");
      }
      for (int j = i; j >= 1; j--) {
        System.out.print(j == i ? j : " " + j);
      }
      System.out.println();
    }
  }

  public static void main(String[] args) {
    Scanner scanner = new Scanner(System.in);

    System.out.print("Please enter an integer between 1 and 9 inclusive: ");
    int userNum = -1;
    while (scanner.hasNext()) {
      if (scanner.hasNextInt()) {
        userNum = scanner.nextInt();
        if (userNum >= 1 && userNum <= 9) {
          scanner.close();
          break;
        } else {
          System.out.println("ERROR: Input number was not between 1 and 9");
          System.out.print("Enter a single digit number: ");
        }
      } else {
        System.out.println("ERROR: Invalid Input");
        System.out.print("Please enter an integer between 1 and 9 inclusive: ");
        scanner.next();
      }
    }

    printPatternTwo(userNum);
  }
}

Answer 4

打印数字时，请在数字之间留一个空格。

// this works up n == 9
System.out.print(" " + j);

对于n> 9，问题是您需要在固定数量的空间上打印数字和空格，但是数字的字符数包含更改。一种解决方案是使用制表符（如果想要很大的数字，则使用制表符）。

// when printing blank spaces use a tab instead
for (int k= userInput; k>=i; k-- ) {
    System.out.printf("\t");
}

// then print number with tab in front
for(int j =i; j>=1; j-- ) {
    System.out.print("\t"+j);
}

Answer 5

如果您对流感到满意：

public class PatternTwo {

  public static void main(String[] args) {
    Scanner in = new Scanner(System.in);
    int userInput;
    System.out.println("Please enter a number 1...9 : ");
    userInput = in.nextInt();
    String s = "";
    int temp = userInput;

    revRange(1, temp).forEach(n -> printNumbers(n, temp));
  }

  static IntStream revRange(int from, int to) {
    return IntStream.range(from, to)
      .map(i -> to - i + from - 1);
  }

  private static void printNumbers(int n, int max) {
    IntStream.rangeClosed(0, n -1)
      .forEach(num -> System.out.print("  "));
    revRange(1, max - n)
      .forEach(num -> System.out.print(num + " "));
    System.out.println();
  }
}

Answer 6

另一种替代方法是使用printf分隔每行的间距并对每个数字使用String.format。

String.format("%1$" + length + "s", inputString)

如果inputString> = inputString.length返回length，否则在inputString中填充空格，使长度等于给定的长度。

示例：

public static void main(String[] args) {
    Scanner in = new Scanner(System.in);
    int userInput;
    System.out.println("Please enter a number 1...9 : ");// you can use this also for numbers > 9
    userInput = in.nextInt();
    int digits = String.valueOf(userInput).length()+1;
    for(int i = 1; i<=userInput; i++){
        System.out.printf("%1$"+(digits*(userInput-i+1))+"s","");
        for(int k = userInput - (userInput-i); k >= 1; k--){
            System.out.print(String.format("%1$"+digits+"s", k));
        }
        System.out.println();
    }
}

如何用Java将此数字表打印到控制台？

6 个答案: