[Java self-study notes # #] the tenth day of basic Java programming structure

Textbook: "Java core technology Volume"

The basic program structure design of Java Chapter III

3.10.8 irregular array

Exercises in the book there is a number k is drawn from inside the number n, calculate your probability of winning.

In general, if the number k extracted from the n numbers, there will be:

n × (n-1) × (n-2) × (n-3) × ...... × (n-k + 1) / 1 × 2 × 3 × ...... × k species possibility.

Since k is always not larger than n, so all of these possibilities in the array, it would be a triangular array, as follows:

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

1 6 15 20 15 6 1 

1, according to a conventional solution sleeve formula as follows:

package array;

public class IrregularArray 
{
	public static void main(String[] args)
	{
		final int NMAX = 10;
		int [] [] odds = new int [NMAX + 1] []; // create a two-dimensional array
		for (int n=0;n<=NMAX;n++)
		{
			odds [n] = new int [n + 1]; // triangular frame structures
		}
		
		for(int n=0;n<odds.length;n++)
		{
			for(int k=0;k<odds[n].length;k++)
			{
				int lotterOdds=1;
				for(int i=1;i<=k;i++)
				{
					lotterOdds = lotterOdds*(n-i+1)/i;
					odds [n] [k] = lotterOdds; // data array into one
				}
				
			}
		}
		for(int[] row:odds)
		{
			for(int odd:row)
				System.out.printf ( "% 4d", odd); // iterate all the elements of the output array
			System.out.println();            
		}
	}

}

　　The results were as follows:

   0
   0   1
   0   2   1
   0   3   3   1
   0   4   6   4   1
   0   5  10  10   5   1
   0   6  15  20  15   6   1
   0   7  21  35  35  21   7   1
   0   8  28  56  70  56  28   8   1
   0   9  36  84 126 126  84  36   9   1
   0  10  45 120 210 252 210 120  45  10   1

2, there is a more tricky, no method of solving equation sets:

Observation triangular array of topics given in advance, the array elements that can satisfy a rule:

When the row index (i) and column index (j) are greater than 0, numbers [i] [j] = numbers [i-1] [j-1] + numbers [i-1] [j]. // Suppose Numbers [] [] array is triangular in question

So the code can also write:

package array;

public  class IrregularArray
{
    public static void main(String[] args)
    {
        Final  int ROWS =. 7 ;
         Final  int colums =. 7 ;
         int [] [] = trianglearrays new new  int [ROWS] [colums]; // create a two-dimensional array of size has been given
         for ( int R & lt = 0; R & lt <ROWS; ++ r )
        {
            trianglearrays [R & lt] [ 0] =. 1 ; // the elements of the first row of the array of charged
        }
        for (int c=0;c<COLUMS;c++)
        {
            trianglearrays [C] [C] =. 1 ; // index equal to the ranks of elements of the array loaded
        }
        for(int c1=1;c1<trianglearrays[0].length;c1++)
        {
            for (int r1=1;r1<trianglearrays.length;r1++)
            {
                if(c1!=r1)
                {
                    trianglearrays [R1] [C1] = trianglearrays [R1-1] [C1-1] + trianglearrays [R1-1 ] [C1]; // set of mathematical laws charged with other elements
                }
            }
        }
        
        for (int [] row:trianglearrays)
        {
            for (int b:row)
                System.out.printf ( "% 4D" , B); // traverse the array element output
            System.out.println();
        }
        
    }

}

The results were as follows:

1 0 0 0 0 0 0
1 1 0 0 0 0 0
1 2 1 0 0 0 0
1 3 3 1 0 0 0
1 4 6 4 1 0 0
1 5 10 10 5 1 0
1 6 15 20 15 6 1

3, ignoring the system automatically filled in the space 0, then two kinds of programming are desirable. The first way, the realization set up a triangular frame approach really new. This also shows that: write code does not just do the math problems the law, but also with coding tools, data structures, and other features, the code seeks to show the best results.