Diamond Challenge

Look at this URL, and see if you can figure out the diamond problem:

http://www.cs.wcu.edu/cscontest/questions.txt

I've already got it figured out, but I would like to see if anyone else can get this, and if their solution looks like mine.

So if we get this your gonna buy a diamond for each of us? :lol:

Too tired to format. :sick:

Code:

---

import java.util.*;

 public class Diamond{
  public static void main(String[] args) {
  try{
    if(args.length < 1)return;
    String s = args[0];
    int k = Integer.parseInt(s);
    for(int i = 1; i <= k ; ++i){
    System.out.println();
    char[] c = new char[i];
    Arrays.fill(c, '*');
    for(int x = 0; x < c.length; ++x){
    System.out.print(c[x]);
    }
  }
    for(int i = k; i >= 0 ; --i){
    System.out.println();
    char[] c = new char[i];
    Arrays.fill(c, '*');
    for(int x = 0; x < c.length; ++x){
    System.out.print(c[x]);
    }
  }
  }catch(NumberFormatException nfe){System.err.println("Must be a valid number!");}
 }
}

---

Might not be good though but here goes.

VB Code:

public class test2{
    public static void main(String[] args){
        int n=Integer.parseInt(args[0]);
        if(n%2==0) return;
        char[][] c=new char[n][n];
        for(int i=0;i<n;i++)
            for(int j=0;j<n;j++) c[i][j]=' ';
        int startPos=(n/2),endPos=(n/2);
        for(int i=0;i<n;i++){
            for(int j=startPos;j<=endPos&&j>=0&&j<n;j++)
                c[i][j]='*';
            startPos=(i<(n/2))?startPos-1:startPos+1;
            endPos=(i<(n/2))?endPos+1:endPos-1;
        }
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                System.out.print(c[i][j]);
            }
            System.out.println();
        }
    }
}

Quote:

---

Originally Posted by Dilenger4

Too tired to format. :sick:

Code:

---

import java.util.*;

 public class Diamond{
  public static void main(String[] args) {
  try{
    if(args.length < 1)return;
    String s = args[0];
    int k = Integer.parseInt(s);
    for(int i = 1; i <= k ; ++i){
    System.out.println();
    char[] c = new char[i];
    Arrays.fill(c, '*');
    for(int x = 0; x < c.length; ++x){
    System.out.print(c[x]);
    }
  }
    for(int i = k; i >= 0 ; --i){
    System.out.println();
    char[] c = new char[i];
    Arrays.fill(c, '*');
    for(int x = 0; x < c.length; ++x){
    System.out.print(c[x]);
    }
  }
  }catch(NumberFormatException nfe){System.err.println("Must be a valid number!");}
 }
}

---

---

Well, you got the easy part done, but were is the spacing.....muuuuhahhah..That's the hardest part!

Quote:

---

Originally Posted by nebulom

Might not be good though but here goes.

VB Code:

public class test2{
    public static void main(String[] args){
        int n=Integer.parseInt(args[0]);
        if(n%2==0) return;
        char[][] c=new char[n][n];
        for(int i=0;i<n;i++)
            for(int j=0;j<n;j++) c[i][j]=' ';
        int startPos=(n/2),endPos=(n/2);
        for(int i=0;i<n;i++){
            for(int j=startPos;j<=endPos&&j>=0&&j<n;j++)
                c[i][j]='*';
            startPos=(i<(n/2))?startPos-1:startPos+1;
            endPos=(i<(n/2))?endPos+1:endPos-1;
        }
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                System.out.print(c[i][j]);
            }
            System.out.println();
        }
    }
}

---

Nice job neb, you got it...Except you have the same problem as me: The spacing inbetween the astrix.

Here was my solution, it wasn't nearly as efficent as neb, but that's ok..

NOTE: Watch and learn dillenger! ;)

Code:

---

import java.io.*;

public class Diamonds
{
        public static int width;
        public static void main(String[] args) throws IOException
        {
                BufferedReader inputReader = new BufferedReader(new InputStreamReader(System.in));
                System.out.print("Enter the width of the largest segment of the triangle:  ");
                width = Integer.parseInt(inputReader.readLine());
                int countDown = 0;
                int counter2 = width-1;
                int count = width-2;

                for (int i=0; i<width; i++)
                {
                        calculateTopSpaces(counter2);
                        for (int j=counter2; j<width; j++)
                        {
                               
                                System.out.print("*");
                               
                        }
                        System.out.println("");
                        counter2 -= 2;
                        if (counter2 < 0)
                        {
                                for (int a=0; a<width; a++)
                                {
                                        calculateBottomSpaces(count);
                                        for (int d=0; d<count; d++)
                                        {
                                               
                                                System.out.print("*");
                                        }
                                        count-=2;
                                        System.out.println("");
                                }
                                break;
                               
                        }
                       
                }       
        }
       
        public static void calculateTopSpaces(int count)
        {
                int spaces = count/2+2;
                for (int i=0; i<spaces; i++)
                {
                        if (i == spaces-1)
                        {
                        }
                        else
                        {
                                System.out.print(" ");
                        }
                }
       
        }
        public static void calculateBottomSpaces(int count)
        {
                        int spaces = (width-count)/2+2;
                for (int i=0; i<spaces; i++)
                {
                        if (i == spaces-1)
                        {
                        }
                        else
                        {
                                System.out.print(" ");
                        }
                }
        }
                       
               
}

---

Is there another one that you guys want to try? I'm practicing those because I have a team competition at that university, and those questions are previous questions....So pick one out!

What about the Pig Latin program? I was having some trouble on that one.

Ill try that one but i probably won't have it done till sat night. It's better for me to save it for the weekend. Could do it friday but that's beer and poker night. :)

Quote:

---

Originally Posted by Dilenger4

Ill try that one but i probably won't have it done till sat night. It's better for me to save it for the weekend. Could do it friday but that's beer and poker night. :)

---

Well, it doesn't matter when it gets done..I mean, anything will help me. I was having trouble with it, and would like to see a solution.

Quote:

---

Originally Posted by Dilenger4

beer and poker night. :)

---

That's a holy time for a man.

Diamond updated

VB Code:

public class test3{
    public static void main(String[] args){
        int n=Integer.parseInt(args[0]);
        if(n%2==0) return;
        n=(n*2)+3;
        char[][] c=new char[n][n];
        for(int i=0;i<n;i++)
            for(int j=0;j<n;j++) c[i][j]=' ';
        int startPos=(n/2),endPos=(n/2);
        for(int i=0;i<n;i+=2){
            for(int j=startPos;j<=endPos&&j>=0&&j<n;j+=2)
                c[i][j]='*';
            startPos=(i<(n/2))?startPos-2:startPos+2;
            endPos=(i<(n/2))?endPos+2:endPos-2;
        }
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                System.out.print(c[i][j]);
            }
            System.out.println();
        }
    }
}

Quote:

---

Originally Posted by nebulom

Diamond updated

VB Code:

public class test3{
    public static void main(String[] args){
        int n=Integer.parseInt(args[0]);
        if(n%2==0) return;
        n=(n*2)+3;
        char[][] c=new char[n][n];
        for(int i=0;i<n;i++)
            for(int j=0;j<n;j++) c[i][j]=' ';
        int startPos=(n/2),endPos=(n/2);
        for(int i=0;i<n;i+=2){
            for(int j=startPos;j<=endPos&&j>=0&&j<n;j+=2)
                c[i][j]='*';
            startPos=(i<(n/2))?startPos-2:startPos+2;
            endPos=(i<(n/2))?endPos+2:endPos-2;
        }
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                System.out.print(c[i][j]);
            }
            System.out.println();
        }
    }
}

---

That is very nice, where are you actually puting the spaces between?

Quote:

---

Originally Posted by System_Error

That is very nice, where are you actually puting the spaces between?

---

Quote:

---

Originally Posted by nebulom

Diamond updated

VB Code:

public class test3{
    public static void main(String[] args){
        int n=Integer.parseInt(args[0]);
        if(n%2==0) return;
        n=(n*2)+3;
        char[][] c=new char[n][n];
        for(int i=0;i<n;i++)
            [hl=#FFFF00]for(int j=0;j<n;j++) c[i][j]=' ';[/hl]
        int startPos=(n/2),endPos=(n/2);
        for(int i=0;i<n;i+=2){
            for(int j=startPos;j<=endPos&&j>=0&&j<n;j+=2)
                c[i][j]='*';
            startPos=(i<(n/2))?startPos-2:startPos+2;
            endPos=(i<(n/2))?endPos+2:endPos-2;
        }
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                System.out.print(c[i][j]);
            }
            System.out.println();
        }
    }
}

---

Premade? :)

I finished the PigLatin program, it was a lot easier than I thought once I started using StringTokenizer.

Code:

---

import java.io.*;
import java.util.*;

public class PigLatin4
{
  public static void main(String[] args) throws Exception
  {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        System.out.print("Enter a phrase --> ");
        String phrase = br.readLine();
        StringTokenizer st = new StringTokenizer(phrase);
       
        while (st.hasMoreTokens())
        {
                StringBuffer sb = new StringBuffer(st.nextToken());
                sb.append(sb.charAt(0));
                sb.append("ay");
                sb.deleteCharAt(0);
                System.out.print(sb.toString() + " ");
        }
  }
}

---