quick and dirty solution using for loops over arrays and bitset

master
Lothar Buchholz 5 years ago
parent 863825ad1a
commit 3034ee26a5

@ -1,6 +1,4 @@
import java.util.*;
import java.io.*;
import java.math.*;
/**
* Template code to help you parse the standard input
@ -10,11 +8,64 @@ class Solution {
public static void main( String[] args ) {
Scanner in = new Scanner( System.in );
// read values with in.next...() methods
// code your solution here
int[][] grid = new int[9][9];
for ( int index = 0; in.hasNext(); index++ ) {
int row = index / 9;
int col = index % 9;
grid[row][col] = in.nextShort();
}
System.out.println( checkHorizontally( grid )
&& checkVertically( grid )
&& checkAllBlocks( grid )
);
}
private static boolean checkHorizontally( int[][] grid ) {
//noinspection ForLoopReplaceableByForEach
for ( int r = 0; r < grid.length; r++ ) {
BitSet bs = new BitSet( 9 );
for ( int c = 0; c < grid[r].length; c++ ) {
bs.set( grid[r][c] );
}
if ( bs.cardinality() != 9 ) return false;
}
return true;
}
private static boolean checkVertically( int[][] grid ) {
for ( int c = 0; c < 9; c++ ) {
BitSet bs = new BitSet( 9 );
//noinspection ForLoopReplaceableByForEach
for ( int r = 0; r < grid.length; r++ ) {
bs.set( grid[r][c] );
}
if ( bs.cardinality() != 9 ) return false;
}
return true;
}
private static boolean checkAllBlocks( int[][] grid ) {
return checkBlock( grid, 0, 0 )
&& checkBlock( grid, 0, 3 )
&& checkBlock( grid, 0, 6 )
&& checkBlock( grid, 3, 0 )
&& checkBlock( grid, 3, 3 )
&& checkBlock( grid, 3, 6 )
&& checkBlock( grid, 6, 0 )
&& checkBlock( grid, 6, 3 )
&& checkBlock( grid, 6, 6 );
}
// Write result with System.out.println()
System.out.println( "value" );
private static boolean checkBlock( int[][] block, int indexTop, int indexLeft ) {
BitSet bs = new BitSet( 9 );
for ( int r = indexTop; r < indexTop + 3; r++ ) {
for ( int c = indexLeft; c < indexLeft + 3; c++ ) {
bs.set( block[r][c] );
}
}
return ( bs.cardinality() == 9 );
}
}

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