[CTCI 4th Edition] 19.2

Description: Design an algorithm to check if some one has won a tic-tac-toe game.

Thought: I think the solution CTCI gives is not clear and understandable. Try the following one.

Code: 

public class TripleT {
    enum State{Blank, X, O};
    int n = 3;
    State[][] board = new State[n][n];
    int moveCount;

    void Move(int x, int y, State s){ // someone puts in (x, y)
     if(board[x][y] == State.Blank) board[x][y] = s;
     moveCount++;

     for(int i = 0; i < n; i++){ // check row
      if(board[x][i] != s) break;
      if(i == n-1)  //report win for s
     }

     for(int i = 0; i < n; i++){ // check col
      if(board[i][y] != s) break;
      if(i == n-1)  //report win for s
     }

     if(x == y){  //check diag
      for(int i = 0; i < n; i++){
       if(board[i][i] != s) break;
       if(i == n-1)  //report win for s
      }
     }

        if(x + y == n - 1){  //check anti diag
             for(int i = 0; i < n; i++){
                if(board[i][(n-1)-i] != s) break;
              if(i == n-1)  //report win for s
      }
     }

 
     //check draw
     if(moveCount == (n^2 - 1)) //report draw

    }
}

Thought: There should also be a O(1) solution for every new move. 

public class TripleT {
    enum State{Blank, X, O};
    int n = 3;
    State[][] board = new State[n][n];
    int moveCount;
    int[] result = new int[2 * n + 2];

    void Move(int x, int y, State s){ // someone puts in (x, y)
    
     moveCount++;
        if (s == State.X) {
            result[x]++;
            if (result[x] == n) //report win for X
            result[n + y]++;
            if (result[n + y] == n) //report win for X
            if (x == y) result[2 * n]++;
            if (result[2 * n] == n) //report win for X
            if (x == n - 1 - y) result[2 * n + 1]++;
            if (result[2 * n + 1] == n) //report win for X
        }
        
        if (s == State.O) {
            result[x]--;
            if (result[x] == -n) //report win for O
            result[n + y]--;
            if (result[n + y] == -n) //report win for O
            if (x == y) result[2 * n]--;
            if (result[2 * n] == -n) //report win for O
            if (x == n - 1 - y) result[2 * n + 1]--;
            if (result[2 * n + 1] == -n) //report win for O
        }
 
     //check draw
     if(moveCount == (n^2 - 1)) //report draw

    }
}