348. Design Tic-Tac-Toe

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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.

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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?

解法1： O(N) move

基本的思路就是每次move之后，检查行，列，对角线是否满足胜利条件，如果满足就返回。

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class TicTacToe {
    /** Initialize your data structure here. */
    int[][] board = null;
    int n;
    public TicTacToe(int n) {
        board = new int[n][n];        
        this.n = 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) {
        board[row][col] = player;
        if (isWin(row, col, player)) return player;
        return 0;
    }
    private boolean isWin(int row, int col, int player) {
        // check row
        boolean win = true;
        for (int i = 0; i < n; i++) {
            if (board[row][i] != player) {
                win = false;
                break;
            }
        }
        if (win) return true;
        win = true;
        // check col
        for (int i = 0; i < n; i++) {
            if (board[i][col] != player) {
                win = false;
                break;
            }
        }
        if (win) return true;
        // check diagonal
        if (row == col) {
            win =  true;
            for (int i = 0; i < n; i++) {
                if (board[i][i] != player) {
                    win = false;
                    break;
                }
            }
        }
        if (win) return true;
        if (row + col == n - 1) {
            win = true;
            for (int i = 0; i < n; i++) {
                if (board[i][n - 1 - i] != player) {
                    win = false;
                    break;
                }
            }
        }
        if (win) return true;
        return false;
    }
}
/**
 * Your TicTacToe object will be instantiated and called as such:
 * TicTacToe obj = new TicTacToe(n);
 * int param_1 = obj.move(row,col,player);
 */

解法2： O(1) move

实际上这题可以做到o(1)的复杂度。用+1和-1表示每一个方向两个选手的落子情况。那么如果一个行都被其中一个选手占领，他的绝对值就一定和行数相等。由此

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class TicTacToe {
    private int[] rows;
    private int[] cols;
    private int diagonal;
    private int antiDiagonal;
    /** Initialize your data structure here. */
    public TicTacToe(int n) {
        rows = new int[n];
        cols = new int[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 toAdd = player == 1 ? 1 : -1;
        rows[row] += toAdd;
        cols[col] += toAdd;
        if (row == col)
        {
            diagonal += toAdd;
        }
        if (col == (cols.length - row - 1))
        {
            antiDiagonal += toAdd;
        }
        int size = rows.length;
        if (Math.abs(rows[row]) == size ||
            Math.abs(cols[col]) == size ||
            Math.abs(diagonal) == size  ||
            Math.abs(antiDiagonal) == size)
        {
            return player;
        }
        return 0;
    }
}
/**
 * Your TicTacToe object will be instantiated and called as such:
 * TicTacToe obj = new TicTacToe(n);
 * int param_1 = obj.move(row,col,player);
 */