ITI0011:uniqueLines

Siin on koodinäide, kuidas võib lahendada teisipäeval, 16. septembril kell 08:00 toimunud praksis tehtud tunniülesannet:

<source lang="java"> /**

* Whether the given one-dimensional array which represents 3x3 table
* has unique numbers in all its rows and columns.
* @author Ago
*
*/

public class T8 {

public static void main(String[] args) { System.out.println(uniqueLines(new int[]{1, 2, 3, 3, 1, 2, 2, 3, 1})); System.out.println(uniqueLines(new int[]{1, 2, 3, 3, 1, 2, 7, 2, 1})); System.out.println(uniqueLinesGeneral(new int[]{1, 2, 3, 3, 1, 2, 2, 3, 1})); System.out.println(uniqueLinesGeneral(new int[]{1, 2, 3, 3, 1, 2, 7, 2, 1})); System.out.println(uniqueLinesGeneral(new int[]{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16})); }

/** * Given an array of numbers, which represents 3x3 table * whether all the rows and columns consist of unique numbers. * @param board One-dimensional array which represents 3x3 table * @return True, if all the columns and rows consist of unique numbers * (not a single row or column consist of two or more of the same numbers) * False, if at least one row or column has at least two elements with * the same value. */ public static boolean uniqueLines(int[] board) { /* * Table with indices: * 0 1 2 * 3 4 5 * 6 7 8 */

// let's just check all 3 rows and columns within one for loop for (int i = 0; i < 3; i++) { // i * 3 will give us the first element in every row: // i = 0: 0 * 3 = 0 // i = 1: 1 * 3 = 3 // i = 2: 2 * 3 = 6

// i * 3 + 1 => second element in the row // i * 3 + 2 => third element in the row

// if first element in a row == second el in a row if (board[i * 3] == board[i * 3 + 1] || // OR second element == third element board[i * 3 + 1] == board[i * 3 + 2] || // OR first element == third element board[i * 3] == board[i * 3 + 2]) { // then we have a duplicate, let's return false // as there is no point to check further return false; }

// i gives us the first element in every column: // 0: first column // 1: second column // 2: third column

// i + 3 => second element in the column // i + 6 => third element in the column

// if the first element in column == second el in a col if (board[i] == board[i + 3] || // OR second element == third element board[i + 3] == board[i + 6] || // OR first element == third element board[i] == board[i + 6]) { // we have a duplicate, return false here return false; } } // if we reach here, there is no duplicate // so we can return true return true; }

/** * General version of uniqueLines which accepts squares with * different size than 3. * @param board One-dimensional array which represents N x N * table. * @return true if every row and column consists of * only unique numbers, false otherwise. * If the input array does not present a square, false * is returned. * @see T8#uniqueLines(int[]) */ public static boolean uniqueLinesGeneral(int[] board) { int size = (int) Math.sqrt(board.length); // let's check whether we have a full square if (size * size != board.length) { System.out.println("Not a square!"); return false; } // below, rows and columns are checked within the same loop // it is basically the same as having separate for loops for rows // and separate loops for columns for (int axis1 = 0; axis1 < size; axis1++) { for (int axis2 = 0; axis2 < size; axis2++) { for (int comp = 0; comp < axis2; comp++) { if (board[axis1 * size + axis2] == board[axis1 * size + comp]) { // axis1 = row // axis2 = column // comp = compare current column value to previous columns' values return false; } if (board[axis2 * size + axis1] == board[comp * size + axis1]) { // axis1 = column // axis2 = row // comp = compare current row value to previous rows' values return false; } } } } // if some row/col had duplicates, we won't make this far // (return ends the execution of this method) // so, if we reach here, there are no duplicates return true; } }

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