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

You may assume the following rules:

  1. A move is guaranteed to be valid and is placed on an empty block.
  2. Once a winning condition is reached, no more moves is allowed.
  3. 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|

分析：http://www.cnblogs.com/grandyang/p/5467118.html
因为要求 “A player who succeeds in placing n of their marks in a horizontal, vertical, or diagonal row wins the game”， 所以我们可以建立一个大小为n的一维数组rows和cols，还有变量对角线diag和逆对角线rev_diag，这种方法的思路是，如果玩家1在
第一行某一列放了一个子，那么rows[0]自增1，如果玩家2在第一行某一列放了一个子，则rows[0]自减1，那么只有当rows[0]等于n或者-n的时候，表示第一行的子都是一个玩家放的，则游戏结束返回该玩家即可，其他各行各列，对角线和逆对角线都是这种思路，

 1 public class TicTacToe {
 2     private int[] rows;
 3     private int[] cols;
 4     private int diag;
 5     private int rev_diag;
 6     private int n;
 7  
 8     /** Initialize your data structure here. */
 9     public TicTacToe(int n) {
10         this.n = n;
11         rows = new int[n];
12         cols = new int[n];
13     }
14      
15     public int move(int row, int col, int player) {
16         int add = player == 1 ? 1 : -1;
17          
18         rows[row] += add;
19         cols[col] += add;
20          
21         diag += (row == col ? add : 0);
22         rev_diag += (row + col == n - 1 ? add : 0);
23              
24         return (Math.abs(rows[row]) == n || Math.abs(cols[col]) == n || Math.abs(diag) == n || Math.abs(rev_diag) == n) ? player : 0;
25     }
26 }