CODEWITHSUNDEEP
javaloops
CSpadawan
CSpadawan

Reputation: 115

Printing a number triangle

CS student here. I've just received an introduction to loops and I'm not sure I understand them very well. I'm trying to print a triangle of numbers n, such that if n = 4 you'd get something like this:

         4
      3  7
   2  6  9
1  5  8 10

Instead I'm winding up with something like:

   4
3   5

Suffice it to say I'm lost. Here's my code:

void drawT3 (int n)
{
    int k = 1;
    int t = 1;
    for (int i=1;i<=n;i++)
    {  
        k = n;
        int j;
        for (j=1;j<=n-i;j++)
            System.out.print(" ");

        for (j=1;j<=t;j++)
        {
            System.out.printf("%3d",k);
            k += (n - j);
        }
        n--;
        t++;
        System.out.println();
    }
}

Upvotes: 4

Views: 1372

Answers (4)

Gaurav Jain
Gaurav Jain

Reputation: 1855

int n=4,i,j,k,t;
for (i=n;i>=1;i--)
{  
    t=i;
    k=n;
    for(j=1;j<i;j++)
        System.out.printf("   ");  // for leading spaces

    System.out.printf("%3d",i);   // for first digit(or number) in each row (in your example these are 4,3,2,1)

    for(j=i;j<n;j++)
    {
        t+=k;   
        System.out.printf("%3d",t);
        k--;
    }
    System.out.print("\n");
}

OUTPUT: for n=8

                      8
                   7 15
                6 14 21
             5 13 20 26
          4 12 19 25 30
       3 11 18 24 29 33
    2 10 17 23 28 32 35
 1  9 16 22 27 31 34 36

http://ideone.com/C1O1GS

make space around numbers according to your need.

PS: I would never suggest to write any pattern code using array unless it is very complicated. array will use extra memory space.

Upvotes: 1

ChuckCottrill
ChuckCottrill

Reputation: 4444

Observe that there are many ways to print out a triangle of numbers as described above, For example, here are two,

// for n=5,
// 1  2  3  4  5
//    6  7  8  9
//      10 11 12
//         13 14
//            15

And

//             5
//          4  9
//       3  8 12
//    2  7 11 14
// 1  6 10 13 15

And since recursion is Fun!

class triangle
{
    //Use recursion,
    static int rowUR( int count, int start, int depth )
    {
        int ndx;
        if(count<=0) return start;
        //-depth?
        for (ndx=0;ndx<depth;ndx++)
        {
            System.out.print("   ");
        }
        //how many? 5-depth, 5,4,3,2,1
        for( ndx=0; ndx<count; ++ndx )
        {
            System.out.printf("%3d",start+ndx);
        }
        System.out.printf("\n");
        if( count>0 )
        {
            rowUR( count-1, ndx+start, depth+1 );
        }
        return ndx;
    }
    //Use recursion,
    static int rowLR( int count, int start, int depth )
    {
        int ndx, accum;
        if( start < count )
            rowLR( count, start+1, depth+1 );
        for( ndx=0; ndx<depth; ++ndx )
        {
            System.out.print("   ");
        }
        accum=start;
        //how many? 5-depth, 1,2,3,4,5
        for( ndx=0; ndx<(count-depth); ++ndx )
        {
            System.out.printf("%3d",accum);
            accum+=count-ndx;
        }
        System.out.printf("\n");
        return ndx;
    }
    public static void main(String[] args)
    {
        int count=4, depth=0, start=1;
        System.out.printf("rowUR\n");
        rowUR( count=5, start=1, depth=0 );
        System.out.printf("rowLL\n");
        rowLL( count=5, start=1, depth=0 );
    }
};

Upvotes: 1

JackTools.Net
JackTools.Net

Reputation: 734

public static void main(String[] args) {
    // TODO code application logic here
    triangle(4);
}

static public void triangle(int n){
    int x = 0;
    for (int i = n;i>0;i--){
        System.out.print(i + " ");
        x = i+n;
        for (int j=0;j<n-i;j++){
            System.out.print(x - j + " ");
            x = x + n -j;
        }
        System.out.println("");
    }
}

Output for 4:

4

3 7

2 6 9

1 5 8 10

Output for 6:

6

5 11

4 10 15

3 9 14 18

2 8 13 17 20

1 7 12 16 19 21

Upvotes: 1

Adam Stelmaszczyk
Adam Stelmaszczyk

Reputation: 19837

void printTriangle(int n)
{ 
    // build an auxiliary 2D array
    final int t[][] = new int[n][n];
    int i = 1;
    for (int s = n - 1; s <= 2 * (n - 1); s++)
    {
        for (int x = s - n + 1; x < n; x++)
        {
            t[x][s - x] = i++;
        }
    }
    // print the array
    for (int y = 0; y < n; y++)
    {
        for (int x = 0; x < n; x++)
        {
            if (t[x][y] > 0)
            {
                System.out.printf("%3d", t[x][y]);
            }
            else
            {
                System.out.printf("   ");
            }
        }
        System.out.println(); // start new line
    }
}
  1. Build an auxiliary 2D array of size n.
  2. Put numbers into array as human will do, from 1 to n, following the diagonals. s in the code represents x + y sum. That sum is constant for every diagonal. In the first diagonal (the longest one) sum is equal to n - 1. In the second diagonal sum is 1 more, n. In the last "diagonal" (bottom right corner) the sum is 2 * (n - 1). That's exactly our loop: for (int s = n - 1; s <= 2 * (n - 1); s++). Having the sum and x we can obtain y with simple subtraction, y = s - x.
  3. Print the array. Each cell of array is initialized with 0 (int's default value). So, if a cell has zero, we just print 3 spaces, to preserve the shape of triangle.

PS. My code was written for "educational purposes" :) To show how it can be done, in easy way. It's not optimized for speed nor memory.

Upvotes: 2

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