348 - Design Tic-Tac-Toe

解法一 - player1 + 1,player2 - 1,記錄每個 row, col diag 的正負值

這個解法非常聰明,巧妙地利用了不會有錯誤 input 的性質,捨棄掉位置資訊,只儲存每個 row, col, diag 是否已經被填了三次,實作如下:

class TicTacToe {
public:
    /** Initialize your data structure here. */
    TicTacToe(int n): rows(n, 0), cols(n, 0) {
        size = n;
        diagonal = 0, antidiagonal = 0;
    }

    /** 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. */
    int move(int row, int col, int player) {
        int toAdd = player == 1 ? 1 : -1;

        rows[row] += toAdd;
        cols[col] += toAdd;
        if(row - col == 0) { diagonal += toAdd; }
        if(row + col == size-1) { antidiagonal += toAdd; }

        if(abs(rows[row]) == size ||
           abs(cols[col]) == size ||
           abs(diagonal) == size  ||
           abs(antidiagonal) == size) {
            return player;
        }

        return 0;
    }

private:
    vector<int> rows, cols;
    int diagonal, antidiagonal, size;
};

/**
 * Your TicTacToe object will be instantiated and called as such:
 * TicTacToe* obj = new TicTacToe(n);
 * int param_1 = obj->move(row,col,player);
 */

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