Java nested loops

问题:

Program Description:

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.

Here is what I want as an output:

Here is what I have so far.

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();
        }
    }
}

This only produces the first half of the X, I do not know how to produce the lower half.

回答1:

Try this one:

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();
}

explanation: The first operate on Xaxis coordinates, second for operates on Yaxis. Our task is to cover diagonal. Covering first diagonal is where coordinateX == coordinateY. In code is if(i==j). These are points (1,1), (2,2)...... Second diagonal are points where (x,y)= (20,1),(19,2),(18,3) .... This situation covers second if(i == ySize - j - 1) .

回答2:

You can try that:

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();}

    }
}

回答3:

Though the above solutions work perfectly, I tried to experiment by not using nested for and the soultion is as below. This will have higher performance than using nested for which has a complexity of O(n2) when compared to O(n) in this.

     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(" ");
    }