用Java打印三角形

Question

I'm practicing basic coding exercises and trying to print the following triangle in Java:

*
***
*****
***
*

The following code gives me the results but I feel like there must be a much more elegant solution

    for (int i = 1; i <= 5; i++) {
        if (i % 2 == 1) {
            for (int j = 1; j <= i; j++) {
                System.out.print("*");
            }
            System.out.println("");
        }

    }
    for (int i = 3; i > 0; i--) {
        if (i % 2 == 1) {
        for (int j = 1; j < i + 1; j++) {
            System.out.print("*");
        }
        System.out.println("");
        }
    } 

Can anyone provide some insight into how to make this work in a better way?

Ok, here's some more code that produces the correct result that uses just the two for loops, but it looks even uglier:

    for (int i = 1; i <= 10; i += 2) {
        if (i <= 5) {
            for (int j = 1; j <= i; j++) {
                System.out.print("*");
            }
            System.out.println("");

        }
        else if(i > 5 && i < 8){
            for(int j = i/2; j > 0; j--){
                System.out.print("*");
            }
            System.out.println("");
        }
        else{
            for(int j = 1; j > 0; j--){
                System.out.print("*");
            }
            System.out.println("");
        }
    }

Answer 1

First, you are skipping each 2nd iteration of the loop because you want to increase two steps at once. You can do this by changing the "i++" in your loop to "i += 2" and "i--" to "i -= 2", that will have the same effect and allows you to remove the if inside both loops.

Another improvement would be using a single outer loop and figuring out whether the inner loop should be increasing or decreasing the amount of asterisks. Maybe you can come up with an equation that gives you the amount of asterisks based on the value of i? (I didn't want to solve it completely so you have some exercise left, just comment if you want a full solution)

Updated with a solution that might be considered elegant as you can change the height of the triangle and there is no repetition:

int height = 5;
for (int i = 1; i <= 2 * height; i += 2) {
    int numAsterisks;
    if (i <= height) {
        numAsterisks = i;
    } else {
        numAsterisks = 2 * height - i;
    }

    for (int j = 0; j < numAsterisks; j++) {
        System.out.print("*");
    }
    System.out.println();
}

Answer 2

What about the following?

public void printTriangle(int size) {
    int half = size / 2;
    for (int i = 0; i < size; i++) {
        int stars = 1 + 2 * (i <= half ? i : size - 1 - i);
        char[] a = new char[stars];
        Arrays.fill(a, '*');
        System.out.println(new String(a));
    }
}

Or just a bit more optimized:

public void printTriangle(int size) {
    int half = size / 2;
    char[] a = new char[size];
    Arrays.fill(a, '*');
    for (int i = 0; i < size; i++) {
        int stars = 1 + 2 * (i <= half ? i : size - 1 - i);
        System.out.println(new String(a, 0, stars));
    }
}

Answer 3

for(int i = 0; i < 7; i++) {
    for(int j = 0; j < i; j++) {
        print("*");
    }
    print("\n");
}

This can be another solution to print a regular right triangle...

Answer 4

Here's a different way of looking at the problem. By using an integer array, I can solve lots of shape drawing problems by changing the values in the array.

When solving more difficult problems, you would use model classes instead of simple integers. The idea, however, is the same.

Here's the output.

*
***
*****
***
*

And here's the code:

public class Triangle {

    public static void main(String[] args) {
        int[] heights = {1, 3, 5, 3, 1};

        for (int i = 0; i < heights.length; i++) {
            for (int j = 0; j < heights[i]; j++) {
                System.out.print("*");
            }
            System.out.println("");
        }
    }

}

Answer 5

How about...

int width = 5;
for (int i = 1; i <= width; i+=2){
    System.out.println(String.format("%"+i+"s", "").replaceAll(" ", "*"));
}
for (int i = width-2; i > 0; i-=2){
    System.out.println(String.format("%"+i+"s", "").replaceAll(" ", "*"));
}

Or, even better yet...

int width = 7;
double half = width / 2
for (int i = 0; i < width; i++){
    System.out.println(String.format("%"+((i < half ? i : (width-i-1))*2+1)+"s", "").replaceAll(" ", "*"));
}

Gives

*
***
*****
***
*