Tic-tac-toe is played by two players A and B on a 3 x 3 grid.
Here are the rules of Tic-Tac-Toe:
- Players take turns placing characters into empty squares (" ").
- The first player A always places "X" characters, while the second player B always places "O" characters.
- "X" and "O" characters are always placed into empty squares, never on filled ones.
- The game ends when there are 3 of the same (non-empty) character filling any row, column, or diagonal.
- The game also ends if all squares are non-empty.
- No more moves can be played if the game is over.
Given an array moves
where each element is another array of size 2 corresponding to the row and column of the grid where they mark their respective character in the order in which A and B play.
Return the winner of the game if it exists (A or B), in case the game ends in a draw return "Draw", if there are still movements to play return "Pending".
You can assume that moves
is valid (It follows the rules of Tic-Tac-Toe), the grid is initially empty and A will play first.
Example 1:
Input: moves = [[0,0],[2,0],[1,1],[2,1],[2,2]] Output: "A" Explanation: "A" wins, he always plays first. "X " "X " "X " "X " "X " " " -> " " -> " X " -> " X " -> " X " " " "O " "O " "OO " "OOX"
Example 2:
Input: moves = [[0,0],[1,1],[0,1],[0,2],[1,0],[2,0]] Output: "B" Explanation: "B" wins. "X " "X " "XX " "XXO" "XXO" "XXO" " " -> " O " -> " O " -> " O " -> "XO " -> "XO " " " " " " " " " " " "O "
Example 3:
Input: moves = [[0,0],[1,1],[2,0],[1,0],[1,2],[2,1],[0,1],[0,2],[2,2]] Output: "Draw" Explanation: The game ends in a draw since there are no moves to make. "XXO" "OOX" "XOX"
Example 4:
Input: moves = [[0,0],[1,1]] Output: "Pending" Explanation: The game has not finished yet. "X " " O " " "
Constraints:
1 <= moves.length <= 9
moves[i].length == 2
0 <= moves[i][j] <= 2
- There are no repeated elements on
moves
. moves
follow the rules of tic tac toe.
Related Topics:
Array
// OJ: https://leetcode.com/problems/find-winner-on-a-tic-tac-toe-game/
// Author: github.com/lzl124631x
// Time: O(1)
// Space: O(1)
class Solution {
string check(int A[3][3]) {
for (int i = 0; i < 3; ++i) {
int j = 1;
while (j < 3 && A[i][0] != 0 && A[i][j] == A[i][0]) ++j;
if (j == 3) return A[i][0] == 1 ? "A" : "B";
j = 1;
while (j < 3 && A[0][i] != 0 && A[j][i] == A[0][i]) ++j;
if (j == 3) return A[0][i] == 1 ? "A" : "B";
}
int i = 1;
while (i < 3 && A[0][0] != 0 && A[i][i] == A[0][0]) ++i;
if (i == 3) return A[0][0] == 1 ? "A" : "B";
i = 1;
while (i < 3 && A[0][2] != 0 && A[i][2 - i] == A[0][2]) ++i;
if (i == 3) return A[0][2] == 1 ? "A" : "B";
for (int i = 0; i < 3; ++i) {
for (int j = 0; j < 3; ++j) {
if (A[i][j] == 0) return "Pending";
}
}
return "Draw";
}
public:
string tictactoe(vector<vector<int>>& moves) {
int A[3][3] = {}, ch = 1;
for (auto &m : moves) {
A[m[0]][m[1]] = ch;
ch = ch == 1 ? 2 : 1;
}
return check(A);
}
};
// OJ: https://leetcode.com/problems/find-winner-on-a-tic-tac-toe-game/
// Author: github.com/lzl124631x
// Time: O(1)
// Space: O(1)
class Solution {
public:
string tictactoe(vector<vector<int>>& moves) {
int A[8] = {}, B[8] = {};
for (int i = 0; i < moves.size(); ++i) {
int r = moves[i][0], c = moves[i][1];
if (i % 2 == 0 && (++A[r] == 3 || ++A[3 + c] == 3 || (r == c && ++A[6] == 3) || (r == 2 - c && ++A[7] == 3))) return "A";
if (i % 2 && (++B[r] == 3 || ++B[3 + c] == 3 || (r == c && ++B[6] == 3) || (r == 2 - c && ++B[7] == 3))) return "B";
}
return moves.size() == 9 ? "Draw" : "Pending";
}
};