LeetCode 348. Design Tic-Tac-Toe（井字棋）

原题网址：https://leetcode.com/problems/design-tic-tac-toe/

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|

Follow up:
Could you do better than O(n2) per move() operation?

Hint:

1. Could you trade extra space such that move() operation can be done in O(1)?
2. You need two arrays: int rows[n], int cols[n], plus two variables: diagonal, anti_diagonal.

方法：类似八皇后问题，记录各行各列以及对角线的落子情况。

public class TicTacToe {    private int[][] rows = new int[2][];    private int[][] cols = new int[2][];    private int[] diag = new int[2];    private int[] adiag = new int[2];    private int n;    /** Initialize your data structure here. */    public TicTacToe(int n) {        for(int i=0; i<n; i++) {            rows[0] = new int[n];            rows[1] = new int[n];            cols[0] = new int[n];            cols[1] = new int[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) {        player --;        rows[player][row] ++;        cols[player][col] ++;        if (row == col) diag[player] ++;        if (row + col == n - 1) adiag[player] ++;        for(int i=0; i<2; i++) {            if (diag[i] == n) return i+1;            if (adiag[i] == n) return i+1;            for(int j=0; j<n; j++) {                if (rows[i][j] == n) return i+1;                if (cols[i][j] == n) return i+1;            }        }        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); */

简化：

public class TicTacToe {    private int[][] rows;    private int[][] cols;    private int[] diag = new int[2];    private int[] adiag = new int[2];    private int size;    /** Initialize your data structure here. */    public TicTacToe(int n) {        size = n;        rows = new int[n][2];        cols = new int[n][2];    }        /** 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) {        rows[row][player-1] ++;        if (rows[row][player-1] == size) return player;        cols[col][player-1] ++;        if (cols[col][player-1] == size) return player;        if (row == col) {            diag[player-1] ++;            if (diag[player-1] == size) return player;        }        if (row+col==size-1) {            adiag[player-1] ++;            if (adiag[player-1] == 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); */