1

到目前为止，我有这个，但我不太确定 printPTriangle 如何使用代码打印三角形。如果有人可以帮助我解决这个问题，我将不胜感激。

public static int factorial(int n) {
    if (n == 1) {
        return 1;
    }
    return n * (factorial(n - 1));
}

public static int pascalsNumber(int x, int y) {
    return factorial(x)/(factorial(y) * factorial((x - y))); //Using combinations formula
}

public static void printPTriangle(int z) {



}
4

4 回答 4

1

尝试这个，

public class PascalTriangle {

public static void main(String[] args) {

    int rows = 10;


    for(int i =0;i<rows;i++) {
        int number = 1;
        System.out.format("%"+(rows-i)*2+"s","");
        for(int j=0;j<=i;j++) {
             System.out.format("%4d",number);
             number = number * (i - j) / (j + 1);

        }
        System.out.println();
    }

}
}

请注意上面用于创建格式良好的三角形的格式化命令。%4d 指示格式化程序在 4 个空格内打印数字。

于 2014-11-05T05:31:02.623 回答
1

我也是 Java 的初学者，正在研究 Pascals 三角形。我喜欢上面指出的格式并将其引入我的代码。我将间距设置得稍大一点，以适应更大的三角形。

导入 java.util.Scanner；

public class Pascal {

    public static void main(String[] args) {

        Scanner scanner = new Scanner(System.in);
        System.out.print("Enter the row number up to which Pascal's triangle has to be printed: ");
        int row = scanner.nextInt();
        print(row);

        scanner.close();
    }

    public static void print(int n) {
        for (int i = 0; i < n; i++) {
            System.out.format("%"+(n-i)*3+"s","");
            for (int j = 0; j <= i; j++) {
                System.out.format("%6d",(pascal(i, j)));
            }
            System.out.println();
        }
    }

    public static int pascal(int i, int j) {
        if (j == 0) {
            return 1;
        } else if (j == i) {
            return 1;
        } else {
            return pascal(i - 1, j - 1) + pascal(i - 1, j);
        }

    }

}
于 2016-02-03T22:01:15.470 回答
0

这是解决方案：

public static void main(String[] args) 
{
    pascal(1,10);
}
static int pascal(int start, int end)
{
    if(start>=end)
        return 0;
    int number = 1;
    System.out.format("%"+(end-start)*2+"s","");
    pascal2(start,number,0);
    System.out.println();
    return pascal(start+1,end);
}
static int pascal2(int start,int number,int end)
{
    if(end>start)
        return 1;
    System.out.format("%4d",number);
    return pascal2(start,number * (start - end) / (end + 1),end+1);
}
于 2017-10-17T19:11:23.990 回答
0

我只是对我的 Java 类的这个编程挑战感到困惑，并且在任何只使用没有循环的递归的地方都找不到帮助。因此，经过数小时的痛苦，它就在这里。它只需要一个主方法或演示类来创建此类的实例PascalsTriangle并使用行数对其进行初始化。

public class PascalsTriangle {

private StringBuilder str; // StringBuilder to display triangle

/**
 * Starts the process of printing the Pascals Triangle
 * @param rows Number of rows to print
 */
public PascalsTriangle(int rows) {

    str = new StringBuilder();

    printTriangle(rows, str);
}

/**
 * Uses recursion to function as an "outer loop" and calls
 * itself once for each row in triangle. Then displays the result
 * @param row The number of the row to generate
 * @param str StringBuilder to insert each row into
 */
public static void printTriangle(int row, StringBuilder str) {

    // calls itself until row equals -1
    if (row >= 0) {

        // calls lower function to generate row and inserts the result into front of StringBuilder
        str.insert(0, getRow(row, 0) + "\n");

        // calls itself with a decremented row number
        printTriangle(row - 1, str);
    } else {

        // when the base case is reached - display the result
        JOptionPane.showMessageDialog(null, str);
        System.exit(0);
    }
}

/**
 * Uses recursion to act as the "inner loop" and calculate each number in the given row
 * @param rowNumber Number of the row being generated
 * @param elementNumber Number of the element within the row (always starts with 0)
 * @return String containing full row of numbers or empty string when base case is reached
 */
public static String getRow(int rowNumber, int elementNumber) {

    // calls itself until elementNumber is greater than rowNumber
    if (elementNumber <= rowNumber) {

        // calculates element using combinations formula: n!/r!(n-r)!
        int element = fact(rowNumber) / (fact(elementNumber) * (fact(rowNumber - elementNumber)));

        // calls itself for each element in row and returns full String            
        return element + " " + getRow(rowNumber, elementNumber + 1);

    } else return "";
}

/**
 * Helper function that uses recursion to calculate factorial of given integer
 * @param n Number to calculate factorial
 * @return Factorial
 */
public static int fact(int n) {
    if (n <= 0)
        return 1;
    else
        return n * fact(n - 1);
}
于 2020-04-02T19:13:53.087 回答