帕斯卡三角形正确格式化java

Question

所以我目前正在处理一项我似乎无法完成的任务。好吧，我已经完成了一切，但想要额外的功劳。我一直在网上四处寻找，但似乎无法找到我正在寻找的确切内容。

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

    System.out.println("Welcome to the Pascal's Triangle program!");
    System.out.println("Please enter the size of the triangle you want");

    int size = kb.nextInt();

    int[][] myArray = new int[size][size];

    myArray = fillArray(myArray); 

    //myArray = calculateArray(myArray);

    printArray(myArray); //prints the array

}

private static int[][] fillArray(int[][] array)
{
    array[0][1] = 1;

    for (int i = 1; i < array.length; i++)
    {
        for (int j = 1; j < array[i].length; j++)
        {
            array[i][j] = array[i-1][j-1] + array[i-1][j];
        }
    }

    return array;
}

private static void printArray(int[][] array)
{
    for (int i = 0; i < array.length; i++)
    {
        for (int j = 0; j < array[i].length; j++)
        {
            if(array[i][j] != 0)
            System.out.print(array[i][j] + " ");
        }
        System.out.println();
    }

}

}

我现在唯一遇到的问题是正确格式化输出，使其看起来像一个实际的三角形。此时任何建议都会非常有帮助。提前致谢

Answer 1

解决此问题的一种方法是，假设您将所有数字格式化为相同的宽度，将问题视为使线条居中的问题。

Java 编码作为练习留给读者，但本质上是：

for lineText : triange lines 
   leadingSpacesCount = (80/2) - lineText.length(); 
   print " " x leadingSpacesCount + lineText

Answer 2

尝试使用http://www.kodejava.org/examples/16.html 的技术创建一个带有array.length - i - 1 空格的数组（需要在数字之间添加数字空格.. 和2 个数字，如果有的话..）。

在外部 for 循环的开头打印此数组。

Answer 3

这里的挑战是您想从三角形的顶部开始打印，但是在到达三角形的最后（也是最宽的）行之前，您不知道将每行居中的位置。诀窍是在您知道最后一行有多宽之前不打印任何内容。一种方法是将所有行生成为String（或StringBuilder）对象并计算最大宽度。然后，从顶部开始，通过首先打印适当数量的空格来使每行居中。正确的空格数为

(maxLineLength - currentLine.length()) / 2

或者，您可以简单地假设最大线条长度并将所有线条居中在该宽度中。如果较长的线超过最大宽度，则三角形将在某一行下方扭曲。 （请确保不要尝试打印负数的空格！）

Answer 4

如果有人正在寻找执行此操作的实际代码，请查看我在 Java 中的实现，它类似于 Craig Taylor 提到的（格式化为相同宽度的数字）加上它使用一种算法来计算没有内存的元素（或阶乘）。

代码有 cmets 解释每个步骤（计算和打印）：

/**
 * This method will print the first # levels of the Pascal's triangle. It
 * uses the method described in:
 * 
 * https://en.wikipedia.org/wiki/Pascal%27s_triangle#Calculating_a_row_or_diagonal_by_itself
 * 
 * It basically computes the Combinations of the current row and col
 * multiplied by the previous one (which will always be 1 at the beginning
 * of each pascal triangle row). It will print each tree element to the output
 * stream, aligning the numbers with spaces to form a perfect triangle.
 * 
 * @param num
 *            # of levels to print
 */
public static void printPascalTriangle(int num) {
    // Create a pad (# of spaces) to display between numbers to keep things
    // in order. This should be bigger than the # of digits of the highest
    // expected number and it should be an odd number (to have the same
    // number of spaces to the left and to the right between numbers)
    int pad = 7;

    // Calculate the # of spaces to the left of each number plus itself
    // (this is the width of the steps of the triangle)
    int stepsWidth = pad / 2 + 1;

    // Now calculate the maximum # of spaces from the left side of the
    // screen to the first triangle's level (we will have num-1 steps in the
    // triangle)
    int spaces = (num - 1) * stepsWidth;

    for (int n = 0; n < num; n++) {
        // Print the left spaces of the current level, deduct the size of a
        // number in each row
        if (spaces > 0) {
            System.out.printf("%" + spaces + "s", "");
            spaces -= stepsWidth;
        }
        // This will represent the previous combination C(n k-1)
        int prevCombination = 1;
        for (int k = 1; k <= n + 1; k++) {
            System.out.print(prevCombination);

            // Calculate how many digits this number has and deduct that to
            // the pad between numbers to keep everything aligned
            int digits = (int) Math.log10(prevCombination);
            if (digits < pad) {
                System.out.printf("%" + (pad - digits) + "s", "");
            }

            // Formula from Wikipedia (we can remove that "+1" if we start
            // the row loop at n=1)
            prevCombination = prevCombination * (n + 1 - k) / k;

        }
        // Row separator
        System.out.println();
    }
}

希望它对某人有所帮助！