Medium
Given a Tic-Tac-Toe board as a string array board
, return true
if and only if it is possible to reach this board position during the course of a valid tic-tac-toe game.
The board is a 3 x 3
array that consists of characters ' '
, 'X'
, and 'O'
. The ' '
character represents an empty square.
Here are the rules of Tic-Tac-Toe:
' '
.'X'
characters, while the second player always places 'O'
characters.'X'
and 'O'
characters are always placed into empty squares, never filled ones.Example 1:
Input: board = [“O “,” “,” “]
Output: false
Explanation: The first player always plays “X”.
Example 2:
Input: board = [“XOX”,” X “,” “]
Output: false
Explanation: Players take turns making moves.
Example 3:
Input: board = [“XOX”,”O O”,”XOX”]
Output: true
Constraints:
board.length == 3
board[i].length == 3
board[i][j]
is either 'X'
, 'O'
, or ' '
.public class Solution {
public boolean validTicTacToe(String[] board) {
// X=1,O=-1,’ ’=0
int sum = 0;
int[] winsCol = new int[3];
int[] winsDiag = new int[2];
boolean xWin = false;
boolean oWin = false;
for (int i = 0; i < 3; i++) {
String str = board[i];
int rowSum = 0;
for (int j = 0; j < 3; j++) {
// char chr=str.toCharArray()[j];
int intchr = 0;
if (str.toCharArray()[j] == 'X') {
intchr = 1;
}
if (str.toCharArray()[j] == 'O') {
intchr = -1;
}
rowSum += intchr;
winsCol[j] += intchr;
if (i == 2 && winsCol[j] == 3) {
xWin = true;
}
if (i == 2 && winsCol[j] == -3) {
oWin = true;
}
if (Math.abs(i - j) != 1) {
if (i == j && i == 1) {
winsDiag[0] += intchr;
winsDiag[1] += intchr;
} else if (i == j) {
winsDiag[0] += intchr;
} else {
winsDiag[1] += intchr;
}
}
if (i == 2 && Math.max(winsDiag[0], winsDiag[1]) == 3) {
xWin = true;
}
if (i == 2 && Math.min(winsDiag[0], winsDiag[1]) == -3) {
oWin = true;
}
}
if (rowSum == 3) {
xWin = true;
}
if (rowSum == -3) {
oWin = true;
}
sum += rowSum;
}
if (sum == 0 && !xWin) {
return true;
}
return sum == 1 && !oWin;
}
}