[LeetCode] 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 player with id player plays at the cell (row, col) of the board. The move is guaranteed to be a valid move.

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

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

设计井字棋。

请在 n × n 的棋盘上，实现一个判定井字棋（Tic-Tac-Toe）胜负的神器，判断每一次玩家落子后，是否有胜出的玩家。

在这个井字棋游戏中，会有 2 名玩家，他们将轮流在棋盘上放置自己的棋子。

在实现这个判定器的过程中，你可以假设以下这些规则一定成立：

      1. 每一步棋都是在棋盘内的，并且只能被放置在一个空的格子里；

      2. 一旦游戏中有一名玩家胜出的话，游戏将不能再继续；

      3. 一个玩家如果在同一行、同一列或者同一斜对角线上都放置了自己的棋子，那么他便获得胜利。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/design-tic-tac-toe
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

这是一道设计题。题目的 follow up 问能不能对于 move() 函数，给出优于 O(n^2) 的思路，我这里给出一个最优解。这个游戏给的棋盘是一个长度为 N 的正方形，只有两个玩家，如果两个玩家有任何一个玩家的棋子摆满了某一行，某一列或者某一条对角线，则判定为胜利，那么我们可以给两个玩家分别赋值 1 和 -1，来判断到底是谁走棋。如果某一行/某一列被同一个玩家占满，则那一行/那一列的sum会是 N 或者 -N。对于对角线也是一样，只是这里对角线的计算有些复杂，但是注意因为是正方形所以计算相对简单

• 正对角线diagonal - 只要判断是否 row == col 即可
• 反对角线antiDiagonal - 判断是否 col == N - row - 1即可

时间O(1)

空间O(n^2)

Java实现

class TicTacToe {
    private int[] rows;
    private int[] cols;
    private int diagonal;
    private int antiDiagonal;

    /** Initialize your data structure here. */
    public TicTacToe(int 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 toAdd = player == 1 ? 1 : -1;
        rows[row] += toAdd;
        cols[col] += toAdd;
        if (row == col) {
            diagonal += toAdd;
        }
        if (col == (cols.length - row - 1)) {
            antiDiagonal += toAdd;
        }

        int size = rows.length;
        if (Math.abs(rows[row]) == size || Math.abs(cols[col]) == size || Math.abs(diagonal) == size
                || Math.abs(antiDiagonal) == size) {
            return player;
        }
        return 0;
    }
}

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

LeetCode 题目总结