(未完成！)LeetCode 348. Design Tic-Tac-Toe 设计井字棋游戏

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|

Follow up:

Could you do better than O(n2) per move() operation?

Hint:

Could you trade extra space such that move() operation can be done in O(1)?

You need two arrays: int rows
, int cols
, plus two variables: diagonal, anti_diagonal.