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How to increase efficiency by using for loops?

I am a beginner java programmer and I am making a simple TicTacToe game using 2D arrays and these are my if statements to check if player 1, or player 2 has won. I believe that this can be simplified by using for loop(s) however I do not understand how to use that method.
if ((grid[0][0] == 1 && grid[0][1] == 1 && grid[0][2] == 1)
|| (grid[1][0] == 1 && grid[1][1] == 1 && grid[1][2] == 1)
|| (grid[2][0] == 1 && grid[2][1] == 1 && grid[2][2] == 1)
|| (grid[0][0] == 1 && grid[1][1] == 1 && grid[2][2] == 1)
|| (grid[0][2] == 1 && grid[1][1] == 1 && grid[2][0] == 1)
|| (grid[0][0] == 1 && grid[1][0] == 1 && grid[2][0] == 1)
|| (grid[0][1] == 1 && grid[1][1] == 1 && grid[2][1] == 1)
|| (grid[0][2] == 1 && grid[1][2] == 1 && grid[2][2] == 1)
&& won == false) {
title.setText("X wins!");
won = true;
} else if ((grid[0][0] == 2 && grid[0][1] == 2 && grid[0][2] == 2)
|| (grid[1][0] == 2 && grid[1][1] == 2 && grid[1][2] == 2)
|| (grid[2][0] == 2 && grid[2][1] == 2 && grid[2][2] == 2)
|| (grid[0][0] == 2 && grid[1][1] == 2 && grid[2][2] == 2)
|| (grid[0][2] == 2 && grid[1][1] == 2 && grid[2][0] == 2)
|| (grid[0][0] == 2 && grid[1][0] == 2 && grid[2][0] == 2)
|| (grid[0][1] == 2 && grid[1][1] == 2 && grid[2][1] == 2)
|| (grid[0][2] == 2 && grid[1][2] == 2 && grid[2][2] == 2)
&& won == false) {
title.setText("O wins!");
won = true;
}
Below is the modified code that uses far less if statments and conditions.
public static boolean hasWon(int[][] grid) {
for (int a = 1; a <= 2; a++) {
for (int b = 0; b < grid.length; b++) {
// Checking for win in horizontal, then vertical, then diagonal
if (grid[b][0] == a && grid[b][1] == a && grid[b][2] == a) {
won = true;
} else if (grid[0][b] == a && grid[1][b] == a && grid[2][b] == a) {
won = true;
} else if ((grid[0][0] == a && grid[1][1] == a && grid[2][2] == a
|| (grid[0][2] == a && grid[1][1] == a && grid[2][0] == a))) {
won = true;
}
}
}
}

In order to help you reach a solution on your own I'll give you some hints for now.
Hint #1: Think about what it means to win. A player must get 3 of their tokens in a row - horizontal, vertical, or diagonal. Think about how that can be represented in your program.
Hint #2: Think about how you can break the problem into smaller more manageable pieces. Think about what each winning scenario has in common and separate that logic into a method that you can call multiple times.
Hint #3: Consider what makes each winning scenario unique and how you might use your grid to produce a representation of the spaces you want to examine that is easier to check for a win.
If you're not sure about how for loops work or other aspects of the Java Language you can find tutorials on Oracle's site

Yes you are right. For loops are the way to go. Here is one way you could implement it.
public class tictactoe {
public static void main(String[] args) {
int[][] grid = {{1, 2, 1},
{1, 2, 1},
{2, 0, 1}};
boolean won = hasWon(grid);
}
public static boolean hasWon(int[][] grid){
for (int player = 1; player <= 2; player++){
boolean playerWon = false;
for(int i = 0; i < 3; i++){
//Horizonal win
playerWon = (grid[i][0] == player && grid[i][1] == player && grid[i][2] == player) || playerWon;
//Vertical Win
playerWon = (grid[0][i] == player && grid[1][i] == player && grid[i][2] == player) || playerWon;
}
//Diagonal Win
playerWon = (grid[0][0] == player && grid[1][1] == player && grid[2][2] == player) || playerWon;
playerWon = (grid[0][2] == player && grid[1][1] == player && grid[2][0] == player) || playerWon;
if(playerWon){
if(player == 1){
System.out.println("X wins!");
return true;
}
else{
System.out.println("O wins!");
return true;
}
}
}
//neither have won
return false;
}
}

Not a direct answer for the question. (as this is not "check all at once" style)
To simplify,
1. Check when a cell is clicked.
2. Condition depends on the place of a cell which is clicked and who clicked the cell.
3. If someone wins, end the game.
code sample
// Part of codes.(not tested.)
// Each cell has three states (0, 1, or 2)
int player = 1; // (Not written here but) switch this each turn (1 or 2)
// In some place (activity.onCreate() etc)
{
// For on click event(0, 0)
cell_0_0.setOnClickListener(
new View.OnClickListener()
{
#Override
public void onClick(View v)
{
grid[0][0] = player;
final boolean isEnd = checkEnd_0_0();
if (isEnd) {
// Call some function to end the game.
// Calling title.setText() in game end function maybe good.
// (as not needed to write many times.)
if (player == 1) {
title.setText("X wins!");
} else {
title.setText("O wins!");
}
} else {
switchPlayer(); // Not written in this code.
}
}
};
);
...
}
// Call this on cell(0, 0) click event
// Returns true if someone wins.
boolean checkEnd_0_0() {
// Omit checking grid[0][0] is 1 or 2 as it is clear.
// Check horizontal.
if (grid[0][1] == player) {
if (grid[0][2] == player) {
return true; // This is the case shown in question.
}
}
// Check other cases (vertical, diagonal)
...
// No match.
return false;
}

(Java) Tic-Tac-Toe game using 2 dimensional Array

In class, our assignment is to create a two-dimensional array and create a tic-tac-toe game around it. I have everything done except displaying when the whole board is full and the game is a draw. I have tried a few things but I have not found the solution and I need some help... Here is my code:
import java.util.Scanner;
public class TicTacToe {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int row, column;
char player = 'X';
//create 2 dimensional array for tic tac toe board
char[][] board = new char[3][3];
char ch = '1';
for (int i = 0; i < 3; i++){
for (int j = 0; j < 3; j++) {
board[i][j] = ch++;
}
}
displayBoard(board);
while(!winner(board) == true){
//get input for row/column
System.out.println("Enter a row and column (0, 1, or 2); for player " + player + ":");
row = in.nextInt();
column = in.nextInt();
//occupied
while (board[row][column] == 'X' || board[row][column] == 'O') {
System.out.println("This spot is occupied. Please try again");
}
//place the X
board[row][column] = player;
displayBoard(board);
if (winner(board)){
System.out.println("Player " + player + " is the winner!");
}
//time to swap players after each go.
if (player == 'O') {
player = 'X';
}
else {
player = 'O';
}
if (winner(board) == false) {
System.out.println("The game is a draw. Please try again.");
}
}
private static void displayBoard(char[][] board) {
for (int i = 0; i < board.length; i++) {
for (int j = 0; j < board[i].length; j++) {
if (j == board[i].length - 1) System.out.print(board[i][j]);
else System.out.print( board[i][j] + " | ");
}
System.out.println();
}
}
//method to determine whether there is an x or an o in the spot
public static Boolean winner(char[][] board){
for (int i = 0; i< board.length; i++) {
for (int j = 0; j < board[0].length; j++) {
if (board[i][j] == 'O' || board[i][j] == 'X') {
return false;
}
}
}
return (board[0][0] == board [0][1] && board[0][0] == board [0][2]) ||
(board[0][0] == board [1][1] && board[0][0] == board [2][2]) ||
(board[0][0] == board [1][0] && board[0][0] == board [2][0]) ||
(board[2][0] == board [2][1] && board[2][0] == board [2][2]) ||
(board[2][0] == board [1][1] && board[0][0] == board [0][2]) ||
(board[0][2] == board [1][2] && board[0][2] == board [2][2]) ||
(board[0][1] == board [1][1] && board[0][1] == board [2][1]) ||
(board[1][0] == board [1][1] && board[1][0] == board [1][2]);
}
}
I want output saying that the board is full when it's full but I get nothing. This is the last line of my output and as you can see, my current strategy is not working as it continues to ask for input. -->
Enter a row and column (0, 1, or 2); for player X:
2 0
X | O | X
O | O | X
X | X | O
Enter a row and column (0, 1, or 2); for player O:

First off:
while (board[row][column] == 'X' || board[row][column] == 'O') {
System.out.println("This spot is occupied. Please try again");
}
This will create a infinite loop because row and column shouldn't change you should ask for new input!
Also
public static Boolean winner(char[][] board){
for (int i = 0; i< board.length; i++) {
for (int j = 0; j < board[0].length; j++) {
if (board[i][j] == 'O' || board[i][j] == 'X') {
return false;
}
}
}
As soon you hit 'O' or 'X' you will exit the Method with a false (no winner)
What you probably want to check is if every spot is occupied
public static Boolean winner(char[][] board){
//Boolean which is true until there is a empty spot
boolean occupied = true;
//loop and check if there is empty space or if its a draw
for (int i = 0; i< board.length; i++) {
for (int j = 0; j < board[0].length; j++) {
//Check if spot is not 'O' or not 'X' => empty
if (board[i][j] != 'O' || board[i][j] != 'X') {
occupied = false;
}
}
}
if(occupied)
return false;
//Check if someone won
return (board[0][0] == board [0][1] && board[0][0] == board [0][2]) ||
(board[0][0] == board [1][1] && board[0][0] == board [2][2]) ||
(board[0][0] == board [1][0] && board[0][0] == board [2][0]) ||
(board[2][0] == board [2][1] && board[2][0] == board [2][2]) ||
(board[2][0] == board [1][1] && board[0][0] == board [0][2]) ||
(board[0][2] == board [1][2] && board[0][2] == board [2][2]) ||
(board[0][1] == board [1][1] && board[0][1] == board [2][1]) ||
(board[1][0] == board [1][1] && board[1][0] == board [1][2]);
}
This would now check if there is a winner or its a tie
Occupied == true == tie == return false
Winner == return true
But you have three states:
Win
Tie
NotFinished
With the changed Method you will NOT finish the game until you win.
Reason:
while(!winner(board) == true)
This makes the game run as long as there is NO winner
(winner() will be false because everything is occupied or there is no winner)
while(!false==true) => while(true)
You could write a method similar to winner but it only checks if the board has empty spots:
public static Boolean hasEmptySpot(char[][] board){
//loop and check if there is empty space
for (int i = 0; i< board.length; i++) {
for (int j = 0; j < board[0].length; j++) {
if (board[i][j] != 'O' && board[i][j] != 'X') {
return true;
}
}
}
return false;
}
//New code
while(hasEmptySpot(board) || !winner(board)){
//Your code for the game here
....
}
this would end the game when there is no empty spot left
After you finished the game you can call winner(board) and it will return if you tied or won!
By creating hasEmptySpot() you could change your winner method to
public static Boolean winner(char[][] board){
return (board[0][0] == board [0][1] && board[0][0] == board [0][2]) ||
(board[0][0] == board [1][1] && board[0][0] == board [2][2]) ||
(board[0][0] == board [1][0] && board[0][0] == board [2][0]) ||
(board[2][0] == board [2][1] && board[2][0] == board [2][2]) ||
(board[2][0] == board [1][1] && board[0][0] == board [0][2]) ||
(board[0][2] == board [1][2] && board[0][2] == board [2][2]) ||
(board[0][1] == board [1][1] && board[0][1] == board [2][1]) ||
(board[1][0] == board [1][1] && board[1][0] == board [1][2]);
}
Why?
Because you finished the game and you know there are only two possible outcomes Win or Tie.
I hope this helped you a little bit.
EDIT
Had a logic error myself!
First mistake:
you still need to check if there is a winner while the game is running forgot that point!
while(hasEmptySpot(board) || !winner(board)){
}
Now this will quit the game loop when there is a winner or no empty spots is left
Second mistake:
In hasEmptySpot()
if (board[i][j] != 'O' && board[i][j] != 'X') {
return true;
not
if (board[i][j] != 'O' || board[i][j] != 'X') {
return true;
Fixed it in the upper examples.
I'm sorry for the inconvenience!

The most efficient way to do this is to keep a running count of how many spaces have been filled previously and increment that count each time a space is occupied. The board can be considered full when that count reaches 9.
If you're familiar with object-oriented programming, I think you'll find this easier to implement if you wrap your 2D array in a Board class.
Example:
public static class Board {
private char[][] spaces = new char[3][3];
private int numMoves = 0;
public void makeMove(int row, int col, char player) {
if (spaces[row][col] == 'X' || spaces[row][col] == 'O') {
System.out.println("This spot is occupied. Please try again");
} else {
spaces[row][col] = player;
numMoves++;
}
}
public boolean isFull() {
return numMoves == 9;
}
public boolean hasWinner() {
...
}
public void display() {
...
}
}

You could try to incorporate a new method such as the following:
public Boolean boardFull()
{
short count = 0;
for(short i = 0; i < 3; i++){
for(short j = 0; j < 3; j++){
if(board[i][j] == ‘O’ || board[i][j] == ’X’){
count++;
} else {
continue;
}
}
}
if(count == 9){
return true;
} else {
return false;
}
}
You could use an if statement to see if it returns true and then print something out if it does.

Solution
The code that's not working is your winner() method. It is always returning false if there is at least one cell occupied. You could proceed based on the last part of Nordiii's answer.
Extra problems
Cell-checking loop
Your code to check if a cell is occupied is going infinitely. You need to use an 'if' statement instead of a 'while' loop:
if(board[row][column] == 'X' || board[row][column] == 'O'){
System.out.println("This spot is occupied. Please try again");
continue;
}
Your old code got stuck always checking if 1 cell was occupied and it always returned true, which kept the loop alive and flooded your console. The continue statement will exit the current iteration of your other 'while' loop and start a new iteration, thus asking for new input.
Exceptions
Man, that's a lot of uncaught exceptions! If I mess up my input, pow! The whole thing fails. Just put a try block for your input-checking code:
try {
row = in.nextInt();
column = in.nextInt();
// Attempt to place player (an ArrayOutOfBoundsException could be thrown)
if(board[row][column] == 'X' || board[row][column] == 'O'){
System.out.println("This spot is occupied. Please try again");
continue;
}
board[row][column] = player;
} catch(Exception e){
System.out.println("I'm sorry, I didn't get that.");
continue;
}
This attempts to execute the code inside the try statement, and if someone inputs something incorrect, the exception gets 'caught' and a new iteration is created. Genius!

Although there are already some great answers I'd like to post another solution that is more generic in its logic to determine the winner. Currently you've hard-coded your some of the possible winning scenarios when you could write more generic logic for this.
As other answers have pointed out you want a method to check for unoccupied spaces in the board and this will tell you if there is a tie. I have implemented such a method in the code below along with the more generic winner logic.
Note that some methods are public to make it easier to test, they do not necessarily have to remain public.
import java.util.Scanner;
public class TicTacToe {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int row, column;
char player = 'X';
//create 2 dimensional array for tic tac toe board
char[][] board = new char[3][3];
char ch = '1';
for (int i = 0; i < 3; i++){
for (int j = 0; j < 3; j++) {
board[i][j] = ch++;
}
}
displayBoard(board);
while(!winner(board) == true){
//get input for row/column
System.out.println("Enter a row and column (0, 1, or 2); for player " + player + ":");
row = in.nextInt();
column = in.nextInt();
//occupied
while (board[row][column] == 'X' || board[row][column] == 'O') {
System.out.println("This spot is occupied. Please try again");
}
//place the X
board[row][column] = player;
displayBoard(board);
if (winner(board)){
System.out.println("Player " + player + " is the winner!");
}
//time to swap players after each go.
if (player == 'O') {
player = 'X';
}
else {
player = 'O';
}
if (winner(board) == false && !hasFreeSpace(board)) {
System.out.println("The game is a draw. Please try again.");
}
}
//Don't forget to close the scanner.
in.close();
}
public static void displayBoard(char[][] board) {
for (int i = 0; i < board.length; i++) {
for (int j = 0; j < board[i].length; j++) {
if (j == board[i].length - 1) System.out.print(board[i][j]);
else System.out.print( board[i][j] + " | ");
}
System.out.println();
}
}
/**
* Determines whether the board is completely occupied by X and O characters
* #param board the board to search through
* #return true if entire board is populated by X or O, false otherwise.
*/
public static boolean hasFreeSpace(char[][] board){
for (int i = 0; i< board.length; i++) {
for (int j = 0; j < board[0].length; j++) {
if (board[i][j] != 'O' && board[i][j] != 'X') {
return true;
}
}
}
return false;
}
//method to determine whether there is a winner
public static boolean winner(char[][] board){
return isHorizontalWin(board) || isVerticalWin(board) || isDiagonalWin(board);
}
/**
* Determines if there is a winner by checking each row for consecutive
* matching tokens.
* #return true if there is a winner horizontally, false otherwise.
*/
private static boolean isHorizontalWin(char[][] board) {
for(int row = 0; row < board.length; row++){
if(isWin(board[row]))
return true;
}
return false;
}
/**
* Determines whether all of the buttons in the specified array have the
* same text and that the text is not empty string.
* #param lineToProcess an array of buttons representing a line in the grid
* #return true if all buttons in the array have the same non-empty text, false otherwise.
*/
private static boolean isWin(char[] lineToProcess) {
boolean foundWin = true;
char prevChar = '-';
for(char character: lineToProcess) {
if(prevChar == '-')
prevChar = character;
if ('O' != character && 'X' != character) {
foundWin = false;
break;
} else if (prevChar != character) {
foundWin = false;
break;
}
}
return foundWin;
}
/**
* Determines whether there is a winner by checking column for consecutive
* matching tokens.
* #return true if there is a vertical winner, false otherwise.
*/
private static boolean isVerticalWin(char[][] board) {
char[] column = null;
//assuming all rows have same legnth (same number of cols in each row), use first row
for(int col = 0; col < board[0].length; col++){
column = new char[board[0].length];
for(int row = 0; row < column.length; row++){
column[row] = board[row][col];
}
if(isWin(column))
return true;
}
return false;
}
/**
* Determines if there is a winner by checking each diagonal for consecutive
* matching tokens.
* #return true if a diagonal winner exists, false otherwise.
*/
private static boolean isDiagonalWin(char[][] board) {
int row = 0, col = 0;
int cols = board.length;
int rows = board[0].length; //assuming all rows are equal length so just use the first one
//Create a one-dimensional array to represent the diagonal. Use the lesser
// of the rows or columns to set its size. If the grid is rectangular then
// a diagonal will always be the size of the lesser of its two dimensions.
int size = rows < cols ? rows : cols;
char[] diagonal = new char[size];
//Since we know the grid is a square we really could just check one of
// these - either row or col, but I left both in here anyway.
while (row < rows && col < cols) {
diagonal[col] = board[row][col];
row++;
col++;
}
if (isWin(diagonal)) {
return true;
}
row = rows - 1;
col = 0;
diagonal = new char[size];
while (row >=0 && col < cols) {
diagonal[col] = board[row][col];
row--;
col++;
}
return isWin(diagonal);
}
}

How do I use a variable that is declared inside of a for loop outside the loop?

I'm creating ATARI BREAKOUT, using the acm.graphics library and I'm trying to access a "brick" outside of my for loop to delete it. I can't figure out any other way to create the bricks without the for loop. Help?
GRect brick = new GRect(brickwidth, brickheight);
for(j = 1; j <= nrows; j++) {
for(i = 0; i < bricksperrow; i++) {
brick.setLocation(i*(brickwidth + brickSep) + 1, brickoffset + j*(brickheight + brickSep));
if(j == 1 || j == 2) {
brick.setColor(Color.RED);
brick.setFilled(true);
}
else if(j == 3 || j == 4) {
brick.setColor(Color.ORANGE);
brick.setFilled(true);
}
else if(j == 5 || j == 6) {
brick.setColor(Color.YELLOW);
brick.setFilled(true);
}
else if(j == 7 || j == 8) {
brick.setColor(Color.GREEN);
brick.setFilled(true);
}
else if(j == 9 || j == 10) {
brick.setColor(Color.CYAN);
brick.setFilled(true);
}
add(brick);
}
}

I guess you want to create many bricks in for loop.
You are doing it wrong, per iteration you are just changing a position of one brick.
You need to create a new brick per iteration and save its reference into some structure preferably a matrix of [nrows, bricksperrow] dimensions.
Here's how:
GRect[][] bricks = new GRect[nrows][bricksperrow];
for(j = 1; j <= nrows; j++) {
for(i = 0; i < bricksperrow; i++) {
bricks[j - 1][i].setLocation(
i*(brickwidth + brickSep) + 1,
brickoffset + j*(brickheight + brickSep));
if(j == 1 || j == 2) {
brick.setColor(Color.RED);
brick.setFilled(true);
}
else if(j == 3 || j == 4) {
brick.setColor(Color.ORANGE);
brick.setFilled(true);
}
else if(j == 5 || j == 6) {
brick.setColor(Color.YELLOW);
brick.setFilled(true);
}
else if(j == 7 || j == 8) {
brick.setColor(Color.GREEN);
brick.setFilled(true);
}
else if(j == 9 || j == 10) {
brick.setColor(Color.CYAN);
brick.setFilled(true);
}
add(bricks[j - 1][i]);
}
}
This way you can have global matrix of bricks from where you can delete any entry.

String to Char || Char to String

Let suppose I have a button
case R.id.button:
which will do the following functionality:
int position;
String keyInStringForm = et2.getText().toString();
int keyInIntegerForm = Integer.parseInt(keyInStringForm);
String text = et1.getText().toString();
for (int i = 0; i < text.length(); i++) {
if (text.charAt(i) == 'a' || text.charAt(i) == 'A') {
position = 0;
break;
} else if (text.charAt(i) == 'b' || text.charAt(i) == 'B') {
position = 1;
break;
} else if (text.charAt(i) == 'c' || text.charAt(i) == 'C') {
position = 2;
break;
} else if (text.charAt(i) == 'd' || text.charAt(i) == 'D') {
position = 3;
break;
} else if (text.charAt(i) == 'e' || text.charAt(i) == 'E') {
position = 4;
break;
} else if (text.charAt(i) == 'f' || text.charAt(i) == 'F') {
position = 5;
break;
} else if (text.charAt(i) == 'g' || text.charAt(i) == 'G') {
position = 6;
break;
} else if (text.charAt(i) == 'h' || text.charAt(i) == 'H') {
position = 7;
break;
} else if (text.charAt(i) == 'i' || text.charAt(i) == 'I') {
position = 8;
break;
} else if (text.charAt(i) == 'j' || text.charAt(i) == 'J') {
position = 9;
break;
} else if (text.charAt(i) == 'k' || text.charAt(i) == 'K') {
position = 10;
break;
} else if (text.charAt(i) == 'l' || text.charAt(i) == 'L') {
position = 11;
break;
} else if (text.charAt(i) == 'm' || text.charAt(i) == 'M') {
position = 12;
break;
} else if (text.charAt(i) == 'n' || text.charAt(i) == 'N') {
position = 13;
break;
} else if (text.charAt(i) == 'o' || text.charAt(i) == 'O') {
position = 14;
break;
} else if (text.charAt(i) == 'p' || text.charAt(i) == 'P') {
position = 15;
break;
} else if (text.charAt(i) == 'q' || text.charAt(i) == 'Q') {
position = 16;
break;
} else if (text.charAt(i) == 'r' || text.charAt(i) == 'R') {
position = 17;
break;
} else if (text.charAt(i) == 's' || text.charAt(i) == 'S') {
position = 18;
break;
} else if (text.charAt(i) == 't' || text.charAt(i) == 'T') {
position = 19;
break;
} else if (text.charAt(i) == 'u' || text.charAt(i) == 'U') {
position = 20;
break;
} else if (text.charAt(i) == 'v' || text.charAt(i) == 'V') {
position = 21;
break;
} else if (text.charAt(i) == 'w' || text.charAt(i) == 'W') {
position = 22;
break;
} else if (text.charAt(i) == 'x' || text.charAt(i) == 'X') {
position = 23;
break;
} else if (text.charAt(i) == 'y' || text.charAt(i) == 'Y') {
position = 24;
break;
} else if (text.charAt(i) == 'z' || text.charAt(i) == 'Z') {
position = 25;
break;
} else if (text.charAt(i) == ' ') {
position = 26;
break;
}
int initialResult = position + keyInIntegerForm;
int finalResult = initialResult % 26;
char resultantChar = alphabets[finalResult];
where as "alphabets" is a char array for a-z characters.
} // for
Now there will be more that one "resultantChar", I want those "resultantChar" to be combined together to form a string so I can set it onto a textview.
How do I do that

If I understood you correctly, try do something like this:
StringBuffer result = new StringBuffer();
for (int i = 0; i < text.length(); i++) {
...
char resultantChar = alphabets[finalResult];
result.append(resultantChar);
}
System.out.println(result);

Please, simplify your code using that!
char ch = 'Z';
ch = Character.toLowerCase(ch);
int position = Character.getNumericValue(ch) - Character.getNumericValue('a');
Or, for your case:
char ch = Character.toLowerCase(text.charAt(i));
if (ch >= 'a' && ch <= 'z') {
position = Character.getNumericValue(ch) - Character.getNumericValue('a');
} else if (ch == ' ') {
position = 26;
}

Use http://developer.android.com/reference/java/lang/StringBuilder.html stringbuilder you can append char with stringbuilder

Tic Tac Toe in Java printing error

I've almost got this homework assignment finished, but I'm having some printing errors with the output.
The game is functional, but there's a problem with methods printOBoard and printXBoard where it messes up the game board when a move is made. The problem becomes worse the longer the game goes on. Thanks for your help.
import java.util.Scanner;
public class TicTacToe {
private enum Tiles {
X, O, EMPTY;
}
public static void main(String[] args) {
int i;
int j;
//initialize 2d array of enum types called "board"
Tiles[][] board = new Tiles[3][3];
for (i=0; i<board.length; i++)
for (j=0; j<board.length; j++)
board[i][j] = Tiles.EMPTY;
//print out an empty board as a 2d array, with each tile set as "EMPTY"
printBoard(board);
int row, col;
int countEmpty=0;
//initial countEmpty count, if it's less than 1, the board is full and the game is over.
for (i=0; i<board.length; i++)
for (j=0; j<board.length; j++)
if (board[i][j] == Tiles.EMPTY)
countEmpty++;
while (countEmpty > 0) {
//Player O enters the row coordinate
System.out.println("Player O's turn.");
System.out.println("Player O: Enter row (0, 1, or 2):");
Scanner stdin1 = new Scanner(System.in);
row = stdin1.nextInt();
//Player O enters the column coordinate
System.out.println("Player O: Enter column (0, 1, or 2):");
Scanner stdin2 = new Scanner(System.in);
col = stdin2.nextInt();
//If the tile is empty, it was a valid move, and an 'O' is placed on the spot.
if (board[row][col] == Tiles.EMPTY) {
board[row][col] = Tiles.O;
//MOVE FOR O ********************
printOBoard(board, row, col);
checkWin(board);
if (checkWin(board) == 2)
;
else
break;
}
//If the tile is not empty, it was not a valid move, and Player O is prompted to try again.
else {
System.out.println("Space already occupied. Please choose another.");
System.out.println("Player O's turn.");
//Player 0 enters the row coordinate
System.out.println("Player O: Enter row (0, 1, or 2):");
stdin1 = new Scanner(System.in);
row = stdin1.nextInt();
//Player O enters the column coordinate
System.out.println("Player O: Enter column (0, 1, or 2):");
stdin2 = new Scanner(System.in);
col = stdin2.nextInt();
//ERROR MOVE FOR O********************
board[row][col] = Tiles.O;
printOBoard(board, row, col);
checkWin(board);
if (checkWin(board) == 2)
;
else
break;
}
//Player X enters the row coordinate
System.out.println("Player X's turn.");
System.out.println("Player X: Enter row (0, 1, or 2):");
Scanner stdin3 = new Scanner(System.in);
row = stdin3.nextInt();
//Player X enters the column coordinate
System.out.println("Player X: Enter column (0, 1, or 2):");
Scanner stdin4 = new Scanner(System.in);
col = stdin4.nextInt();
if (board[row][col] == Tiles.EMPTY) {
board[row][col] = Tiles.X;
printXBoard(board, row, col);
//MOVE FOR X *************************************************
checkWin(board);
if (checkWin(board) == 2)
;
else
break;
}
else {
System.out.println("Space already occupied. Please choose another.");
System.out.println("Player O's turn.");
System.out.println("Player O: Enter row (0, 1, or 2):");
stdin3 = new Scanner(System.in);
row = stdin3.nextInt();
//Player O enters the column coordinate
System.out.println("Player O: Enter column (0, 1, or 2):");
stdin4 = new Scanner(System.in);
col = stdin4.nextInt();
board[row][col] = Tiles.O;
//ERROR MOVE FOR X ****************************************
printXBoard(board, row, col);
checkWin(board);
if (checkWin(board) == 2)
;
else
break;
}
//After both players move, we check to see if the board is full.
countEmpty = 0;
for (i=0; i<board.length; i++)
for (j=0; j<board.length; j++)
if (board[i][j] == Tiles.EMPTY)
countEmpty++;
}
}
//method printBoard prints out a grid of EMPTY's and returns nothing
private static void printBoard(Tiles board[][]) {
int i, j;
System.out.println(" -----------------------------");
System.out.println("| | | |");
for (i=0; i<board.length; i++){
for (j=0; j<board.length; j++){
System.out.printf("| " + board[i][j] + " ");
}
System.out.println("|");
System.out.println("| | | |");
System.out.println(" -----------------------------");
if (i<2)
System.out.println("| | | |");
}
return;
}
//method printXBoard prints out the grid modified with the addition of an X after Player X's turn
private static void printXBoard(Tiles board[][], int curRow, int curCol) {
int i, j;
System.out.println(" -----------------------------");
System.out.println("| | | |");
for (i=0; i<board.length; i++){
for (j=0; j<board.length; j++){
if (i == curRow && j == curCol)
board[i][j] = Tiles.X;
else
;
System.out.printf("| " + board[i][j] + " ");
}
System.out.println("|");
System.out.println("| | | |");
System.out.println(" -----------------------------");
if (i<2)
System.out.println("| | | |");
}
return;
}
//method printOBoard prints out the grid modified with the addition of an X after Player X's turn
private static void printOBoard(Tiles board[][], int curRow, int curCol) {
int i, j;
System.out.println(" -----------------------------");
System.out.println("| | | |");
for (i=0; i<board.length; i++){
for (j=0; j<board.length; j++){
if (i == curRow && j == curCol)
board[i][j] = Tiles.O;
else
;
System.out.printf("| " + board[i][j] + " ");
}
System.out.println("|");
System.out.println("| | | |");
System.out.println(" -----------------------------");
if (i<2)
System.out.println("| | | |");
}
return;
}
//method checkWin checks all possible winning combinations for both players.
private static int checkWin(Tiles board[][]) {
if (board[0][0] == Tiles.X && board[0][1] == Tiles.X && board[0][2] == Tiles.X){
System.out.println("Player X wins!");
return 1;
}
else if (board[0][0] == Tiles.X && board[1][1] == Tiles.X && board[2][2] == Tiles.X){
System.out.println("Player X wins!");
return 1;
}
else if (board[0][0] == Tiles.X && board[1][0] == Tiles.X && board[2][0] == Tiles.X){
System.out.println("Player X wins!");
return 1;
}
else if (board[1][0] == Tiles.X && board[1][1] == Tiles.X && board[1][2] == Tiles.X){
System.out.println("Player X wins!");
return 1;
}
else if (board[2][0] == Tiles.X && board[2][1] == Tiles.X && board[2][2] == Tiles.X){
System.out.println("Player X wins!");
return 1;
}
else if (board[0][1] == Tiles.X && board[1][1] == Tiles.X && board[2][1] == Tiles.X){
System.out.println("Player X wins!");
return 1;
}
else if (board[0][2] == Tiles.X && board[1][2] == Tiles.X && board[2][2] == Tiles.X){
System.out.println("Player X wins!");
return 1;
}
else if (board[2][0] == Tiles.X && board[1][1] == Tiles.X && board[0][2] == Tiles.X){
System.out.println("Player X wins!");
return 1;
}
//check if player O wins
else if (board[0][0] == Tiles.O && board[0][1] == Tiles.O && board[0][2] == Tiles.O){
System.out.println("Player O wins!");
return 0;
}
else if (board[0][0] == Tiles.O && board[1][1] == Tiles.O && board[2][2] == Tiles.O){
System.out.println("Player O wins!");
return 0;
}
else if (board[0][0] == Tiles.O && board[1][0] == Tiles.O && board[2][0] == Tiles.O){
System.out.println("Player O wins!");
return 0;
}
else if (board[1][0] == Tiles.O && board[1][1] == Tiles.O && board[1][2] == Tiles.O){
System.out.println("Player O wins!");
return 0;
}
else if (board[2][0] == Tiles.O && board[2][1] == Tiles.O && board[2][2] == Tiles.O) {
System.out.println("Player O wins!");
return 0;
}
else if (board[0][1] == Tiles.O && board[1][1] == Tiles.O && board[2][1] == Tiles.O) {
System.out.println("Player O wins!");
return 0;
}
else if (board[0][2] == Tiles.O && board[1][2] == Tiles.O && board[2][2] == Tiles.O){
System.out.println("Player O wins!");
return 0;
}
else if (board[2][0] == Tiles.O && board[1][1] == Tiles.O && board[0][2] == Tiles.O) {
System.out.println("Player O wins!");
return 0;
}
else
return 2;
}
}

There is a problem in 2nd attempt to place X (if space is occupied by O), you copy pasted code from O and haven't changed player & tile.
System.out.println("Player O's turn.");
board[row][col] = Tiles.O;
It was pain to read this code -_- but you'll get better eventually ;) .

The reason as to why this is happening when printing out the Letters onto your board(which I've run on my mac) is that when you're printing the letters, you're using println...since you're accompanying a certain amount of space per block you should be using printf...for example:
System.out.printf("%10s", LetterVariableForTurn);
This will give the LetterVariableForTurn 10 spaces of text to write within.
here is a link to help you more: printf
Hope this helps!