Difference between revisions of "2D array problem"
| Line 101: | Line 101: | ||
|Solution= | |Solution= | ||
| − | First a quick overview of how we are going to solve | + | First a quick overview of how we are going to solve the problem. The main body is a good place to start. The body reads a lot like the description of the problem. The two for loops scan the array, and at each cell we check the row, column and diagonal. |
<pre> | <pre> | ||
public static void main(String[] args) { | public static void main(String[] args) { | ||
| + | //for each row | ||
for( int row = 0; row < array[0].length; row++ ) | for( int row = 0; row < array[0].length; row++ ) | ||
{ | { | ||
| + | //and each column | ||
for( int col = 0; col < array.length; col++ ) | for( int col = 0; col < array.length; col++ ) | ||
{ | { | ||
| − | checkRow( row, col ); | + | checkRow( row, col ); //check the row starting here |
| − | checkCol( row, col ); | + | checkCol( row, col ); //check the column starting here |
| − | checkDiagonal( row, col ); | + | checkDiagonal( row, col ); //check the diagonal starting here |
} | } | ||
} | } | ||
| + | //print the best numbers we found | ||
printLine( bestNumbers, bestNumber ); | printLine( bestNumbers, bestNumber ); | ||
} | } | ||
| Line 120: | Line 123: | ||
| − | + | Next, filling in the printLine method is a good idea as it is useful in debugging. Given an array and an int this method prints the output like in my example. The main loop here should look a little strange to you. We want to print "x" between the numbers but not at the end. So we will print toPrintArray[ 0 ] first, then add a "x" and the next value in the loop. | |
<pre> | <pre> | ||
//prints a nice output for a provided line | //prints a nice output for a provided line | ||
| Line 137: | Line 140: | ||
</pre> | </pre> | ||
| − | With printing out of the way the next method to write is checkLine. This method checks if a set of numbers is the new highest. The current highest set of number and their product is | + | With printing out of the way the next method to write is checkLine. This method checks if a set of numbers is the new highest. The current highest set of number and their product is stored in bestNumbers and bestNumber. Manually check a few lines to make sure you are getting the right results. |
<pre> | <pre> | ||
static int[] bestNumbers = new int[lineLength]; | static int[] bestNumbers = new int[lineLength]; | ||
| Line 158: | Line 161: | ||
} | } | ||
</pre> | </pre> | ||
| − | |||
| − | |||
| − | |||
Revision as of 02:56, 9 April 2010
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ProblemThis problem is a simplified version of problem 11 on projecteuler.net. Using the provided array find 4 adjacent numbers along any row, column or diagonal with the greatest product. To keep things simple we will only worry about rows going from left-right, columns going top-bottom, and diagonals going down to the right. I have included some skeleton code to make problem a little easier. For example at row 0, column 0 there are three lines to check.
public class matrix {
static final int lineLength = 4; //length of line to check
static int[] bestNumbers = new int[lineLength]; //best line found so far
static int bestNumber = 0; //product of the best line
static final int array[][] ={
{ 8, 2,22,97,38,15, 0,40, 0,75, 4, 5, 7,78,52,12,50,77,91, 8},
{49,49,99,40,17,81,18,57,60,87,17,40,98,43,69,48, 4,56,62, 0},
{81,49,31,73,55,79,14,29,93,71,40,67,53,88,30, 3,49,13,36,65},
{52,70,95,23, 4,60,11,42,69,24,68,56, 1,32,56,71,37, 2,36,91},
{22,31,16,71,51,67,63,89,41,92,36,54,22,40,40,28,66,33,13,80},
{24,47,32,60,99, 3,45, 2,44,75,33,53,78,36,84,20,35,17,12,50},
{32,98,81,28,64,23,67,10,26,38,40,67,59,54,70,66,18,38,64,70},
{67,26,20,68, 2,62,12,20,95,63,94,39,63, 8,40,91,66,49,94,21},
{24,55,58, 5,66,73,99,26,97,17,78,78,96,83,14,88,34,89,63,72},
{21,36,23, 9,75, 0,76,44,20,45,35,14, 0,61,33,97,34,31,33,95},
{78,17,53,28,22,75,31,67,15,94, 3,80, 4,62,16,14, 9,53,56,92},
{16,39, 5,42,96,35,31,47,55,58,88,24, 0,17,54,24,36,29,85,57},
{86,56, 0,48,35,71,89, 7, 5,44,44,37,44,60,21,58,51,54,17,58},
{19,80,81,68, 5,94,47,69,28,73,92,13,86,52,17,77, 4,89,55,40},
{04,52, 8,83,97,35,99,16, 7,97,57,32,16,26,26,79,33,27,98,66},
{88,36,68,87,57,62,20,72, 3,46,33,67,46,55,12,32,63,93,53,69},
{04,42,16,73,38,25,39,11,24,94,72,18, 8,46,29,32,40,62,76,36},
{20,69,36,41,72,30,23,88,34,62,99,69,82,67,59,85,74, 4,36,16},
{20,73,35,29,78,31,90, 1,74,31,49,71,48,86,81,16,23,57, 5,54},
{01,70,54,71,83,51,54,69,16,92,33,48,61,43,52, 1,89,19,67,48} };
public static void main(String[] args) {
for( int row = 0; row < array[0].length; row++ )
{
for( int col = 0; col < array.length; col++ )
{
checkRow( row, col );
checkCol( row, col );
checkDiagonal( row, col );
}
}
printLine( bestNumbers, bestNumber );
}
//prints a nice output for a provided line
public static void printLine( int[] toPrintArray, int toPrintProd )
{
}
//calculate the product for the line passed
//if it the product is better than bestNumber update it
public static void checkLine( int[] toCheck )
{
}
//from (row,col) check product of the row
public static void checkRow( int row, int col )
{
int[] tempRow = new int[ 4 ];
...
checkLine( tempRow );
}
//from (row,col) check product of the column
public static void checkCol( int row, int col )
{
int[] tempRow = new int[ 4 ];
...
checkLine( tempRow );
}
//from (row,col) check product of the diagonal
public static void checkDiagonal( int row, int col )
{
int[] tempRow = new int[ 4 ];
...
checkLine( tempRow );
}
}
|
More with Arrays
| |
|---|---|---|
SolutionFirst a quick overview of how we are going to solve the problem. The main body is a good place to start. The body reads a lot like the description of the problem. The two for loops scan the array, and at each cell we check the row, column and diagonal.
public static void main(String[] args) {
//for each row
for( int row = 0; row < array[0].length; row++ )
{
//and each column
for( int col = 0; col < array.length; col++ )
{
checkRow( row, col ); //check the row starting here
checkCol( row, col ); //check the column starting here
checkDiagonal( row, col ); //check the diagonal starting here
}
}
//print the best numbers we found
printLine( bestNumbers, bestNumber );
}
//prints a nice output for a provided line
//8 x 2 x 22 x 97 = 34144
public static void printLine( int[] toPrintArray, int toPrintProd )
{
System.out.print( toPrintArray[ 0 ] );
for( int i = 1; i < toPrintArray.length; i++ )
{
System.out.print( " x " + toPrintArray[ i ] );
}
System.out.println( " = " + toPrintProd );
}
With printing out of the way the next method to write is checkLine. This method checks if a set of numbers is the new highest. The current highest set of number and their product is stored in bestNumbers and bestNumber. Manually check a few lines to make sure you are getting the right results.
static int[] bestNumbers = new int[lineLength];
static int bestNumber = 0;
public static void checkLine( int[] toCheck )
{
int tempProduct = 1;
for( int i = 0; i < toCheck.length; i++ )
{
tempProduct = tempProduct * toCheck[i];
}
if( tempProduct > bestNumber )
{
bestNumber = tempProduct;
bestNumbers = toCheck;
}
}
//from (row,col) check product of the row
public static void checkRow( int row, int col )
{
//make sure the row we are checking does not exceed the array row length
if( col + lineLength <= array[0].length )
{
int[] tempRow = new int[ 4 ];
for( int i = 0; i < lineLength; i++ )
{
tempRow[ i ] = array[ row ][ col + i ];
}
checkLine( tempRow );
}
}//checkRow
You're almost there, fill in checkCol, checkDiagonal, and verify that you are getting the right answer. | ||
Code | ||
