348. Design Tic-Tac-Toe

题目链接：https://leetcode.com/problems...

这道题找是否有player赢的方法和N-Queens相似，稍微简化了。统计行列和两个对角线player的情况，两个player分别用+1和-1来记。然后判断是否有一个人赢只需要O(1)的复杂度。当然这么做的前提是假设所有的move都是valid的，棋盘一个地方已经被占用了，就不能走那个地方了。

public class TicTacToe {
    int n;
    int[] cols;
    int[] rows;
    int diag;
    int antidiag;
    
    public TicTacToe(int n) {
        this.n = n;
        cols = new int[n];
        rows = new int[n];
        diag = 0;
        antidiag = 0;
    }
    
    public int move(int row, int col, int player) {
        // -1 for player 1, +1 for player 2
        int flag = (player == 1 ? -1 : 1);
        
        rows[row] += flag;
        cols[col] += flag;
        if(row == col) diag += flag;
        if(row + col == n - 1) antidiag += flag;
        
        if(Math.abs(rows[row]) == n || Math.abs(cols[col]) == n || Math.abs(diag) == n || Math.abs(antidiag) == n) return player;
        
        return 0;
    }
}