Member Avatar for deadllama19

There were two other variation which I completed but I am confused with this one:

Write a method for each triangle variation. Use nested for loops to draw the triangles.

Variation 3:

/** Precondition: width is an odd number >1*/
public void diamond (int width)
{

}

if width = 7, the output should be:

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

  • 3 Contributors
  • 2 Replies
  • 110 Views
  • 1 Day Discussion Span
  • Latest Post Latest Post by GTJava

Dani AI

Generated

The sample wants the widest row printed twice, which is why many "standard" diamond solutions (one peak only) won't match. A clear way to guarantee the requested shape is to iterate rows from 0 through width (inclusive) and derive the star count from min(row, width - row). That yields the sequence 1, 3, 5, ..., width, width, ..., 5, 3, 1 and keeps the logic simple.

A compact, nested-loop Java implementation (uses StringBuilder rather than repeated string concatenation):

/** Precondition: width is an odd number > 1 */
public void diamond(int width) {
    for (int row = 0; row <= width; row++) {
        int k = Math.min(row, width - row);
        int stars = 2 * k + 1;
        int spaces = (width - stars) / 2;

        StringBuilder line = new StringBuilder();
        for (int s = 0; s < spaces; s++) line.append(' ');
        for (int t = 0; t < stars; t++) line.append('*');

        System.out.println(line.toString());
    }
}

Troubleshooting notes and variants:

  • If you see one middle row instead of two, your outer loop likely runs row < width instead of <= width, or your star formula uses width - 1 in the wrong places.
  • Use StringBuilder (or prefilled char[]) for performance; repeatedly using += inside loops is slow.
  • To get the more common single-peak diamond, print the top half up to the middle, then print the bottom half starting one row below the middle (or loop row = 0; row < width; row++ and adjust the min formula).
  • 's posted snippet produces the single-peak form; if the exercise requires the duplicate center, change the loop bounds or use the min(row, width-row) approach above. 's suggestion to split responsibilities (compute counts, print a row) is useful for clarity and testing.

Recommended Answers

All 2 Replies

Member Avatar for parry_kulk

Hope this can help you. I found it while surfing

/** Precondition: width is an odd number >1*/
public void diamond (int width)
{
	   int middlePoint = width/2;
	   int currentRow=0;
	   
	   for(int i=0; i<width ;i++){
	   
		   String spaces = "";
		   for(int j=0;j<Math.abs(middlePoint-currentRow);j++){
				spaces+=" ";
		   }
	 
		   String stars="*";
		   int noOfStars;
		   
		   if(currentRow <= middlePoint){
				noOfStars = currentRow;
		   }else{
				noOfStars = width-currentRow-1;
		   }     
			
		   for(int k=0;k<noOfStars;k++){
				stars+="**";      
		   }
						
		   System.out.println(spaces+stars );
		   
		   currentRow++;
	   
	   }
 

}
Member Avatar for GTJava

Try using multiple methods... and make sure your LOGIC makes sense (Write it out and make sure it makes sense before you go into coding). If you post your code I can help.

Be a part of the DaniWeb community

We're a friendly, industry-focused community of developers, IT pros, digital marketers, and technology enthusiasts meeting, networking, learning, and sharing knowledge.