Java 嵌套循环

Question

计划说明：

Write a program to print 21 rows of X's in the shape of a large X as illustrated below. Be sure so the two rows intersect at the "11" row.

这是我想要的输出：

这是我目前的情况。

public class Program168h {

    public static void main (String [] args)  {
        String d= "X";
        for (int a = 1; a < 23; a++) {
            for (int b = a; b >= 1; b--) {   
                System.out.print(" ");
            }
            System.out.print(d);
            for (int x = a; x < 22; x++) {
                System.out.print("  ");
            }
            System.out.print(d);
            System.out.println();
        }
    }
}

这个只生成X的前半部分，下半部分不知道怎么生成。

Answer 1

试试这个：

int xSize = 21;
int ySize = 21;
String sign = "X";

for (int i = 0; i < xSize; ++i) {
    for (int j = 0; j < ySize; ++j) {
        if (i == j) {
            System.out.print(sign);
        } else if (i == ySize - j - 1) {
            System.out.print(sign);
        } else {
            System.out.print(" ");
        }

    }
    System.out.println();
}

说明：第一个对 X 轴坐标进行操作，第二个对 Y 轴进行操作。我们的任务是覆盖对角线。覆盖第一个对角线是 coordinateX == coordinateY。在代码中是 if(i==j)。这些是点 (1,1), (2,2)...... 第二条对角线是点 (x,y)= (20,1),(19,2),(18,3) .. .. 这种情况涵盖第二个 if(i == ySize - j - 1) 。

Answer 2

你可以试试：

public class ProductX {
        public static void main(String[] args) {
        for (int i = 0; i <= 10; i++) {
                for (int j = 0; j < 10; j++) {
                System.out.print(" ");
                if (i == j) {
                    System.out.print("X");
                }
                if(j == 9-i){
                    System.out.print("X");
                }
            }
        System.out.println();}

    }
}

Answer 3

虽然上面的解决方案很完美，但我尝试不使用嵌套 for 进行实验，结果如下。这将比使用嵌套具有更高的性能，其中与 O(n) 相比具有 O(n2) 的复杂性。

     public void testXFormation() {
        final int countOfLines = 21;
        int countOfSpaceBefore = 0;
        int countOfSpacesAfter = countOfLines -2 ;// 2 characters
        boolean halfReached = false;
        for (int index = 0; index < countOfLines; index++) {
            printSpaces(countOfSpaceBefore); // print required no. of spaces
            System.out.print("x"); // print first x
            printSpaces(countOfSpacesAfter); // print required no. of spaces after x
            if (index != (countOfLines / 2))// Avoid printing double, in the middle
                System.out.print("x");
            System.out.println(""); // move to next line

            /* Once you reach half matrix we need to reverse the logic */
            if (index >= (countOfLines - 1) / 2) {
                halfReached = true;
            }

            /* Reversing the logic for the spaces to be printed */
            if (halfReached) {
                countOfSpaceBefore--;
                countOfSpacesAfter += 2;
            } else {
                countOfSpaceBefore++;
                countOfSpacesAfter -= 2;
            }
        }

    }

    private void printSpaces(int count) {
        for (int i = 0; i < count; i++)
            System.out.print(" ");
    }

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