# 348.Design-Tic-Tac-Toe

## 题目地址

https://leetcode.com/problems/design-tic-tac-toe/

## 题目描述

``````Design a Tic-tac-toe game that is played between two players on a n x n grid.

You may assume the following rules:

A move is guaranteed to be valid and is placed on an empty block.
Once a winning condition is reached, no more moves is allowed.
A player who succeeds in placing n of their marks in a horizontal, vertical, or diagonal row wins the game.
Example:
Given n = 3, assume that player 1 is "X" and player 2 is "O" in the board.

TicTacToe toe = new TicTacToe(3);

toe.move(0, 0, 1); -> Returns 0 (no one wins)
|X| | |
| | | |    // Player 1 makes a move at (0, 0).
| | | |

toe.move(0, 2, 2); -> Returns 0 (no one wins)
|X| |O|
| | | |    // Player 2 makes a move at (0, 2).
| | | |

toe.move(2, 2, 1); -> Returns 0 (no one wins)
|X| |O|
| | | |    // Player 1 makes a move at (2, 2).
| | |X|

toe.move(1, 1, 2); -> Returns 0 (no one wins)
|X| |O|
| |O| |    // Player 2 makes a move at (1, 1).
| | |X|

toe.move(2, 0, 1); -> Returns 0 (no one wins)
|X| |O|
| |O| |    // Player 1 makes a move at (2, 0).
|X| |X|

toe.move(1, 0, 2); -> Returns 0 (no one wins)
|X| |O|
|O|O| |    // Player 2 makes a move at (1, 0).
|X| |X|

toe.move(2, 1, 1); -> Returns 1 (player 1 wins)
|X| |O|
|O|O| |    // Player 1 makes a move at (2, 1).
|X|X|X|
Could you do better than O(n2) per move() operation?``````

## 代码

### Approach #1

``````class TicTacToe {
private int[] rows;
private int[] cols;
private int diagonal;
private int antiDiagonal;

public TicTacToe(int n) {
rows = new int[n];
cols = new int[n];
}

public int move(int row, int col, int player) {
int size = rows.length;
int toAdd = (player == 1 ? 1 : -1);
int target = (player == 1 ? size : -size);

if (row == col) {
}

if (col + row == size - 1) {
}

if (rows[row] == target ||
cols[col] == target ||
diagonal == target ||
antiDiagonal == target) {

return player;
}

return 0;
}
}``````

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