348. Design Tic-Tac-Toe

Assume the following rules are for the tic-tac-toe game on an n x n board between two players:

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 are allowed.
3. A player who succeeds in placing n of their marks in a horizontal, vertical, or diagonal row wins the game.

Implement the TicTacToe class:

• TicTacToe(int n) Initializes the object the size of the board n.
• int move(int row, int col, int player) Indicates that the player with id player plays at the cell (row, col) of the board. The move is guaranteed to be a valid move.

 

Example 1:

Input
["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]]
Output
[null, 0, 0, 0, 0, 0, 0, 1]

Explanation
TicTacToe ticTacToe = new TicTacToe(3);
Assume that player 1 is "X" and player 2 is "O" in the board.
ticTacToe.move(0, 0, 1); // return 0 (no one wins)
|X| | |
| | | |    // Player 1 makes a move at (0, 0).
| | | |

ticTacToe.move(0, 2, 2); // return 0 (no one wins)
|X| |O|
| | | |    // Player 2 makes a move at (0, 2).
| | | |

ticTacToe.move(2, 2, 1); // return 0 (no one wins)
|X| |O|
| | | |    // Player 1 makes a move at (2, 2).
| | |X|

ticTacToe.move(1, 1, 2); // return 0 (no one wins)
|X| |O|
| |O| |    // Player 2 makes a move at (1, 1).
| | |X|

ticTacToe.move(2, 0, 1); // return 0 (no one wins)
|X| |O|
| |O| |    // Player 1 makes a move at (2, 0).
|X| |X|

ticTacToe.move(1, 0, 2); // return 0 (no one wins)
|X| |O|
|O|O| |    // Player 2 makes a move at (1, 0).
|X| |X|

ticTacToe.move(2, 1, 1); // return 1 (player 1 wins)
|X| |O|
|O|O| |    // Player 1 makes a move at (2, 1).
|X|X|X|

 

Constraints:

• 2 <= n <= 100
• player is 1 or 2.
• 0 <= row, col < n
• (row, col) are unique for each different call to move.
• At most n2 calls will be made to move.

 

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

以上是题目，这道题我一拿到手，我想了一个非常暴力的解法，就是每次move()的时候，遍历整个board，看是否有人赢了，时间复杂度是O(n2). 但是follow-up问我们是否可以用更好的时间复杂度来解这道题。

实际上，每次move()的时候，只要check当前的player是否赢了即可，就是check 当前player的row, col, row和col对应的Diagonal 和Anti-Diagonal是否全部被当前player占领即可。

时间复杂度是O(n), 空间复杂度是O(n2), 算法如下：

    private int[][] board;
    private int n;
    public TicTacToe(int n) {
        board = new int[n][n];
        this.n = n;
    }
    
    public int move(int row, int col, int player) {
        board[row][col] = player;
        if(checkHorizon(row, player))
            return player;
        if(checkVertical(col, player))
            return player;
        if(checkDiagonal(row, col, player))
            return player;
        if(checkAntiDiagonal(row, col, player))
            return player;
        return 0;
    }
    
    private boolean checkHorizon(int row, int player){
        for(int i=0;i<n;i++){
            if(board[row][i]!=player)
                return false;
        }
        return true;
    }
    
     private boolean checkVertical(int col, int player){
        for(int i=0;i<n;i++){
            if(board[i][col]!=player)
                return false;
        }
         return true;
    }
    
    private boolean checkDiagonal(int row, int col, int player){
        if(row!=col)
            return false;
        for(int i=0;i<n;i++){
            if(board[i][i]!=player)
                return false;
        }
        return true;
    }
    
    private boolean checkAntiDiagonal(int row, int col, int player){
        for(int i=0;i<n;i++){
            if(board[i][n-i-1]!=player)
                return false;
        }
        return true;
    }

上面的算法每次都要重新去数每行，每列，每个斜线是否都被占满，那么能不能利用曾经数过的结果做个叠加呢？这样就不用每次从头数一遍了？答案是：Yes！

其实我们每次move()的时候，只要把每行，每列，每个斜线都累加起来，如果某行，某列或者某斜线的总和等于n，这个player就赢了。但是一共两个player，怎么累加呢？简单，只要用1和-1分别为两个player做累加即可。总和是n，player1赢，总和是-1，player2赢。

时间复杂度：O(1),  空间复杂度O(n), 算法如下：

class TicTacToe {
    int n;
    private int[] rows;
    private int[] cols;
    int diaNum = 0;
    int antiDiaNum = 0;
    public TicTacToe(int n) {
        rows = new int[n];
        cols = new int[n];
        this.n = n;
    }
    
    public int move(int row, int col, int player) {
        int add = player==1?1:-1;
        int count = player==1?n:-n;
        
            rows[row]+=add;
            cols[col]+=add;
            if(row==col){
                diaNum+=add;
            }
            if(row==n-col-1){
                antiDiaNum+=add;
            }
            if(rows[row]==count||cols[col]==count||diaNum==count||antiDiaNum==count)
                return player;
        return 0;
    }
}