348. Design Tic-Tac-Toe设计井字游戏

［抄题］：

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|

 [暴力解法]：

时间分析：

空间分析：

 [优化后]：

时间分析：

空间分析：

［奇葩输出条件］：

［奇葩corner case］：

［思维问题］：

不知道怎么表示数学关系：每一行都不能有2种元素。所以必须要rows[row] == row才行

［英文数据结构或算法，为什么不用别的数据结构或算法］：

［一句话思路］：

［输入量］：空： 正常情况：特大：特小：程序里处理到的特殊情况：异常情况（不合法不合理的输入）：

［画图］：

［一刷］：

1. 对角线元素也不是直接加一的，也要加addon来表示区别
2. 反向相加也行，所以最后的结果是绝对值

［二刷］：

［三刷］：

［四刷］：

［五刷］：

  [五分钟肉眼debug的结果]：

［总结］：

反向相加也行，所以最后的结果是绝对值

［复杂度］：Time complexity: O(n) Space complexity: O(n)

［算法思想：迭代/递归/分治/贪心］：

［关键模板化代码］：

［其他解法］：

［Follow Up］：

［LC给出的题目变变变］：

 [代码风格] ：

 [是否头一次写此类driver funcion的代码] ：

 [潜台词] ：

class TicTacToe {
  private int[] rows;
  private int[] cols;
  private int diagnonal;
  private int antidiagonal;

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

    public int move(int row, int col, int player) {
      //initialization: add 1 or -1
      int addOn = (player == 1) ? 1 : -1;
      
        //handle 4 cases
      rows[row] += addOn;
      cols[col] += addOn;
      
      if (row == col) {
        diagnonal += addOn;
      }
      
      if (row + col == cols.length - 1) {
        antidiagonal += addOn;
      }
      
      //exit if element == row
      int size = rows.length;
      if (Math.abs(rows[row]) == size || Math.abs(cols[col]) == size || Math.abs(diagnonal) == size || Math.abs(antidiagonal) == size)
      {
          return player;
      }
      return 0;
    }
}