Question

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?

Hint:

1. Could you trade extra space such that move() operation can be done in O(1)?
2. You need two arrays: int rows[n], int cols[n], plus two variables: diagonal, anti_diagonal.

Answer

对于move操作，简单粗暴的方法是遍历整个二维数组，对每一行每一列以及对角线检查。时间，空间复杂度均为O(n^2)

我们可以单独考虑行，列，对角。

对于一行row[i]，如果这一行中有棋手1下过，则+1。如果有棋手2下过，则-1。所以如果这一行全为棋手1下的，row[i]=n。棋手1获胜。

Similar case for col and diagonal, anti diagonal。

Time complexity O(1) Space complexity O(n)

 1 public class TicTacToe {
 2 
 3     private int[] row;
 4     private int[] col;
 5     private int diagonal;
 6     private int anti_diagonal;
 7     private int size;
 8     /** Initialize your data structure here. */
 9     public TicTacToe(int n) {
10         size = n;
11         row = new int[n];
12         col = new int[n];
13         Arrays.fill(row, 0);
14         Arrays.fill(col, 0);
15         diagonal = 0;
16         anti_diagonal = 0;
17     }
18     
19     /** Player {player} makes a move at ({row}, {col}).
20         @param row The row of the board.
21         @param col The column of the board.
22         @param player The player, can be either 1 or 2.
23         @return The current winning condition, can be either:
24                 0: No one wins.
25                 1: Player 1 wins.
26                 2: Player 2 wins. */
27     public int move(int row, int col, int player) {
28         int change = player == 1 ? 1 : -1;
29         this.row[row] += change;
30         this.col[col] += change;
31         if (row == col) {
32             diagonal += change;
33         }
34         if (row == (size - col - 1)) {
35             anti_diagonal += change;
36         }
37         if (this.row[row] == size || this.col[col] == size || diagonal == size || anti_diagonal == size) {
38             return 1;
39         }
40         if (this.row[row] == -size || this.col[col] == -size || diagonal == -size || anti_diagonal == -size) {
41             return 2;
42         }
43         return 0;
44     }
45 }
46 
47 /**
48  * Your TicTacToe object will be instantiated and called as such:
49  * TicTacToe obj = new TicTacToe(n);
50  * int param_1 = obj.move(row,col,player);
51  */

Design Tic-Tac-Toe 解答