348.Design-Tic-Tac-Toe

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|
Follow up:
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);

    rows[row] += toAdd;
    cols[col] += toAdd;
    if (row == col) {
      diagonal += toAdd;
    }

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

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

      return player;
    }

    return 0;
  }
}