348. Design Tic-Tac-Toe

提示给的太直白了。。

比如player 1占据了（0,1），那么row[0]++ col[1]++

表示第一行有1个O，第一列有1个X，假设PLAYER 1最终在第一行连成一排，那最终row[0] == n。

player 2占据了(0,2)，那么 row[0]-- col[2]--

如果PLAYER最终在第三列连成一排，那么col[2] == -n

总之选手A，走到(a,b) 那个格所在行列+1，如果是对角线对角线+1，B的话就-1，先到N的就赢了。。。

public class TicTacToe
{

    int[] rows;
    int[] cols;
    int d1;
    int d2;
    int n;

    /** Initialize your data structure here. */
    public TicTacToe(int n)
    {
        this.n = 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 p = 0;
        if(player == 1) p = 1;
        else p = -1;
        rows[row] += p;
        if(rows[row] == n || rows[row] == -n) return player;

        cols[col] += p;
        if(cols[col] == n || cols[col] == -n) return player;

        if(row == col)
        {
            d1 += p;
            if(d1 == n || d1 == -n) return player;
        }

        if(row+col == n-1)
        {
            d2 += p;
            if(d2 == n || d2 == -n) return player;
        }
        return 0;

    }

}

没看提示前上头了啊。。和打DOTA一个德性。。

一开始那个办法太难了，非要做对，做了好久，提示的办法10分钟就搞定了，次奥。。

贴个一开始的办法。。

用x*n + y当做KEY，PLAYER当做VALUE来建立MAP。。

然后每次查MAP看有没有人赢。。检查MAP的时候要以当前点为中点，横纵坐标分别各从-N到+N，对角线同时-N到+N，然后一个-N到+N，一个+N到-N。

好蠢……

public class TicTacToe
{

    Map<Integer,Integer> map;
    int n;
    /** Initialize your data structure here. */
    public TicTacToe(int n)
    {
        this.n = n;
        map = new HashMap<Integer,Integer>();
    }

    /** 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 key = row*n+col;
        map.put(key,player);
        if(check(key,player)) return player;
        else return 0;
    }

    public boolean check(int k,int p)
    {
        // horizontal
        int i = 0;
        int row = k/n;
        int col = k%n;
        boolean[] check = new boolean[n];

        int c = n-1;

        for(int j = -c; j <= c; j++)
        {
            int key = (row+j)*n + col;
            if(row+j < 0 || row + j >= n) i = 0;
            else if(map.containsKey(key) && map.get(key) == p)
            {

                check[i++] = true;
                if(i == n)
                {
                    boolean temp = true;
                    for(int m = 0; m < n;m++)
                    {
                        temp = temp&&check[m];
                        if(!temp) break;
                    }
                    if(temp) return true;
                }

            }
            else i = 0;

        }
        i = 0;

        for(int j = -c; j <= c; j++)
        {
            int key = row*n + col+j;

            if(col+j < 0 || col + j >= n) i = 0;
            else if(map.containsKey(key) && map.get(key) == p)
            {

                check[i++] = true;
                if(i == n)
                {
                    boolean temp = true;
                    for(int m = 0; m < n;m++)
                    {
                        temp = temp&&check[m];
                        if(!temp) break;
                    }
                    if(temp) return true;
                }
            }
            else i = 0;

        }
        i = 0;

        for(int j = -c; j <= c; j++)
        {
            int key = (row+j)*n + col-j;
            if(row + j < 0 || col-j < 0 || row + j >= n || col -j >= n) i = 0;
            else if(map.containsKey(key) && map.get(key) == p)
            {

                check[i++] = true;
                if(i == n)
                {
                    boolean temp = true;
                    for(int m = 0; m < n;m++)
                    {
                        temp = temp&&check[m];
                        if(!temp) break;
                    }
                    if(temp) return true;
                }
            }
            else i = 0;

        }

        i = 0;
        for(int j = -c; j <= c; j++)
        {
            int key = (row-j)*n + col-j;
            if(row - j < 0 || col-j < 0 || row - j >= n || col -j >= n) i = 0;
            else if(map.containsKey(key) && map.get(key) == p)
            {

                check[i++] = true;
                if(i == n)
                {
                    boolean temp = true;
                    for(int m = 0; m < n;m++)
                    {
                        temp = temp&&check[m];
                        if(!temp) break;
                    }
                    if(temp) return true;
                }
            }
            else i = 0;

        }

        return false;

/*
["TicTacToe","move","move","move","move","move","move","move"]
[[3],[0,0,1],[0,2,2],[2,2,1],[1,1,2],[2,0,1],[1,0,2],[2,1,1]]
["TicTacToe","move","move","move"]
[[2],[0,0,2],[1,1,1],[0,1,2]]
["TicTacToe","move","move","move"]
[[2],[0,1,1],[1,1,2],[1,0,1]]

*/

    }
}

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

438MS 被100%的人干死了。。

空间CHEKC[]是N

MAP是N

早看提示就好了。