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:

  1. Could you trade extra space such that move() operation can be done in O(1)?
  2. You need two arrays: int rows[n], int cols[n], plus two variables: diagonal, anti_diagonal.

看了hint之后想到：既然用数组的话，一行一列都是一个element来代表，估计这个element是要用sum了，那么，能不能用sum来代表一行，使它有且只有一种可能，全部是player1完成的/全部是Player2；所以想到了是Player1就+1，是player2就-1，看最后sum是不是n，或者-n；n的情况只有一种情况这一行全是player1。因为说了不会有invalid move, 所以情况是唯一的

 1 public class TicTacToe {
 2     int[] rows;
 3     int[] cols;
 4     int diagonal;
 5     int anti_diagonal;
 6     int size;
 7 
 8     /** Initialize your data structure here. */
 9     public TicTacToe(int n) {
10         rows = new int[n];
11         cols = new int[n];
12         diagonal = 0;
13         anti_diagonal = 0;
14         size = n;
15     }
16     
17     /** Player {player} makes a move at ({row}, {col}).
18         @param row The row of the board.
19         @param col The column of the board.
20         @param player The player, can be either 1 or 2.
21         @return The current winning condition, can be either:
22                 0: No one wins.
23                 1: Player 1 wins.
24                 2: Player 2 wins. */
25     public int move(int row, int col, int player) {
26         int change = (player==1? 1 : -1);
27         rows[row] += change;
28         cols[col] += change;
29         if (row == col) diagonal += change;
30         if (row + col == size-1) anti_diagonal += change;
31         if (Math.abs(rows[row])==size || Math.abs(cols[col])==size || Math.abs(diagonal)==size || Math.abs(anti_diagonal)==size)
32             return player;
33         return 0;
34     }
35 }
36 
37 /**
38  * Your TicTacToe object will be instantiated and called as such:
39  * TicTacToe obj = new TicTacToe(n);
40  * int param_1 = obj.move(row,col,player);
41  */