Algorithm to determine status of a TicTacToe game?

Algorithm to determine status of a TicTacToe game?

I have a program that allows 2 players to play TicTacToe. After each
player makes a move, it should display the board at that point and return
an enumaration called Status that show whether the players should
continue, if a player won, or if it's a draw. However, the algorithm
either return a StackOverflowError, or continues input. Here is the
algorithm I used.
//Checks for winner by rows
for (int i = 0; i < 3; i++) {
for (int j = 0; j < 1; j++) {
if (board[i][j] == 'X') {
if (board[i][j] == board[i][0 + 1] && board[i][j] ==
board[i][0 + 2]) {
printStatus(1);
return Status.WIN;
} else {
return Status.CONTINUE;
}
} else if (board[i][j] == 'O') {
if (board[i][j] == board[i][0 + 1] && board[i][j] ==
board[i][0 + 2]) {
printStatus(2);
return Status.WIN;
} else {
return Status.CONTINUE;
}
}
}
}
//Checks for winner by columns
for (int i = 0; i < 1; i++) {
for (int j = 0; j < 3; j++) {
if (board[i][j] == 'X') {
if (board[i][j] == board[0 + 1][j] && board[i][j] ==
board[0 + 2][j]) {
printStatus(1);
return Status.WIN;
} else {
return Status.CONTINUE;
}
} else if (board[i][j] == 'O') {
if (board[i][j] == board[0 + 1][j] && board[i][j] ==
board[0 + 2][j]) {
printStatus(1);
return Status.WIN;
} else {
return Status.CONTINUE;
}
}
}
}
//This group of if statements boards for winner diagnolly
if (board[0][0] == 'X') {
if (board[0][0] == board[1][1] && board[0][0] == board[2][2]) {
printStatus(1);
return Status.WIN;
} else {
return Status.CONTINUE;
}
}else if (board[0][0] == '0') {
if (board[0][0] == board[1][1] && board[0][0] == board[2][2]) {
printStatus(1);
return Status.WIN;
} else {
return Status.CONTINUE;
}
}
if (board[0][2] == 'O') {
if (board[0][2] == board[1][1] && board[0][2] == board[2][0]) {
printStatus(1);
return Status.WIN;
} else {
return Status.CONTINUE;
}
}else if (board[0][2] == 'X') {
if (board[0][2] == board[1][1] && board[0][2] == board[2][0]) {
printStatus(1);
return Status.WIN;
} else {
return Status.CONTINUE;
}
}