[LeetCode] 348. Design Tic-Tac-Toe

MobService 发布于2019-08-16 13:51 / 3073人阅读

Problem

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:

</>复制代码 
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?

Solution

</>复制代码 
class TicTacToe {
    private int[] rows;
    private int[] cols;
    private int diagonal;
    private int antiDiagonal;
    private int size;
    /** Initialize your data structure here. */
    public TicTacToe(int n) {
        rows = new int[n];
        cols = new int[n];
        diagonal = 0;
        antiDiagonal = 0;
        size = n;
    }
    
    /** Player {player} makes a move at ({row}, {col}).
        @param row The row of the board.
        @param col The column of the board.
        @param player The player, can be either 1 or 2.
        @return The current winning condition, can be either:
                0: No one wins.
                1: Player 1 wins.
                2: Player 2 wins. */
    public int move(int row, int col, int player) {
        int num = player == 1 ? 1 : -1;
        rows[row] += num;
        cols[col] += num;
        if (row == col) diagonal += num;
        if (row == size-1-col) antiDiagonal += num;
        if (Math.abs(rows[row]) == size ||
           Math.abs(cols[col]) == size ||
           Math.abs(diagonal) == size ||
           Math.abs(antiDiagonal) == size) return player;
        return 0;
    }
}

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