0348. Design Tic-Tac-Toe
https://leetcode.com/problems/design-tic-tac-toe
Description
Assume the following rules are for the tic-tac-toe game on an n x n
board between two players:
A move is guaranteed to be valid and is placed on an empty block.
Once a winning condition is reached, no more moves are allowed.
A player who succeeds in placing
n
of their marks in a horizontal, vertical, or diagonal row wins the game.
Implement the TicTacToe
class:
TicTacToe(int n)
Initializes the object the size of the boardn
.int move(int row, int col, int player)
Indicates that the player with idplayer
plays at the cell(row, col)
of the board. The move is guaranteed to be a valid move.
Example 1:
**Input**
["TicTacToe", "move", "move", "move", "move", "move", "move", "move"]
[[3], [0, 0, 1], [0, 2, 2], [2, 2, 1], [1, 1, 2], [2, 0, 1], [1, 0, 2], [2, 1, 1]]
**Output**
[null, 0, 0, 0, 0, 0, 0, 1]
**Explanation**
TicTacToe ticTacToe = new TicTacToe(3);
Assume that player 1 is "X" and player 2 is "O" in the board.
ticTacToe.move(0, 0, 1); // return 0 (no one wins)
|X| | |
| | | | // Player 1 makes a move at (0, 0).
| | | |
ticTacToe.move(0, 2, 2); // return 0 (no one wins)
|X| |O|
| | | | // Player 2 makes a move at (0, 2).
| | | |
ticTacToe.move(2, 2, 1); // return 0 (no one wins)
|X| |O|
| | | | // Player 1 makes a move at (2, 2).
| | |X|
ticTacToe.move(1, 1, 2); // return 0 (no one wins)
|X| |O|
| |O| | // Player 2 makes a move at (1, 1).
| | |X|
ticTacToe.move(2, 0, 1); // return 0 (no one wins)
|X| |O|
| |O| | // Player 1 makes a move at (2, 0).
|X| |X|
ticTacToe.move(1, 0, 2); // return 0 (no one wins)
|X| |O|
|O|O| | // Player 2 makes a move at (1, 0).
|X| |X|
ticTacToe.move(2, 1, 1); // return 1 (player 1 wins)
|X| |O|
|O|O| | // Player 1 makes a move at (2, 1).
|X|X|X|
Constraints:
2 <= n <= 100
player is
1
or2
.0 <= row, col < n
(row, col)
are unique for each different call tomove
.At most
n2
calls will be made tomove
.
Follow-up: Could you do better than O(n2)
per move()
operation?
ac
class TicTacToe {
private int[] r1, r2, c1, c2, d1, d2;
private int n;
/** Initialize your data structure here. */
public TicTacToe(int n) {
this.n = n;
r1 = new int[n];
r2 = new int[n];
c1 = new int[n];
c2 = new int[n];
d1 = new int[2];
d2 = new int[2];
}
/** Player {player} makes a move at ({row}, {col}).
@param row The row of the board.
@param col The column of the board.
@param player The player, can be either 1 or 2.
@return The current winning condition, can be either:
0: No one wins.
1: Player 1 wins.
2: Player 2 wins. */
public int move(int row, int col, int player) {
// compute slope
double mid = (n - 1) / 2.0;
double slope = col == mid ? Double.MAX_VALUE : (row - mid) / (col - mid);
if (player == 1) {
r1[row]++;
c1[col]++;
if (slope == 1) {
d1[0]++;
} else if (slope == -1) {
d1[1]++;
} else if (row == mid && col == mid) {
d1[0]++;
d1[1]++;
}
if (r1[row] == n || c1[col] == n || d1[0] == n || d1[1] == n) return 1;
} else {
r2[row]++;
c2[col]++;
if (slope == 1) {
d2[0]++;
} else if (slope == -1) {
d2[1]++;
} else if (row == mid && col == mid) {
d2[0]++;
d2[1]++;
}
if (r2[row] == n || c2[col] == n || d2[0] == n || d2[1] == n) return 2;
}
return 0;
}
}
/**
* Your TicTacToe object will be instantiated and called as such:
* TicTacToe obj = new TicTacToe(n);
* int param_1 = obj.move(row,col,player);
*/
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