348. 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? 

完全按照题的描述做的 效果出乎意料 beats 99.28%! 但是用的空间复杂度是O(n^2)

每走一这次 判断行列是否被一个player占据 然后判断这个点是否在对角线或反对角线上 再判断整条线是否被占据

代码不太优雅 就不粘出来了

看了discuss 发现了一个666的写法 时间复杂度O(1)... 题目只是要求低于O(n^2)而已

The key observation is that in order to win Tic-Tac-Toe you must have the entire row or column.

public 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;//1.
        
        rows[row] += toAdd;
        cols[col] += toAdd;
        if (row == col)
        {
            diagonal += toAdd;//对角线
        }
        
        if (col == (cols.length - row - 1))
        {
            antiDiagonal += toAdd;//反对角线
        }
        
        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;
    }
}

分别用两个数组rows,cols表示move情况 代替了n*n的二维board

关键代码在1处 把player1,player2分别写作1,-1 每次move 把对应的row和col累加

如果想赢 那么只能是一个player占据了整条线 如果占据了整行 就是row=3或者-3

所以这个思想的奇妙之处 在于他做到了同时统计player1和player2的情况

精髓在于player1,player2是此消彼长的 只有一个占据了所有 才算赢