348 Design Tic-Tac-Toe

348. Design Tic-Tac-Toe​
1. Question
Design a Tic-tac-toe game that is played between two players on anxngrid.
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:
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Given n = 3, assume that player 1 is "X" and player 2 is "O" in the board.
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​
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TicTacToe toe = new TicTacToe(3);
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​
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toe.move(0, 0, 1); -> Returns 0 (no one wins)
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|X| | |
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| | | |    // Player 1 makes a move at (0, 0).
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| | | |
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​
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toe.move(0, 2, 2); -> Returns 0 (no one wins)
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|X| |O|
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| | | |    // Player 2 makes a move at (0, 2).
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| | | |
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​
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toe.move(2, 2, 1); -> Returns 0 (no one wins)
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|X| |O|
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| | | |    // Player 1 makes a move at (2, 2).
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| | |X|
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​
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toe.move(1, 1, 2); -> Returns 0 (no one wins)
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|X| |O|
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| |O| |    // Player 2 makes a move at (1, 1).
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| | |X|
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​
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toe.move(2, 0, 1); -> Returns 0 (no one wins)
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|X| |O|
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| |O| |    // Player 1 makes a move at (2, 0).
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|X| |X|
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​
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toe.move(1, 0, 2); -> Returns 0 (no one wins)
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|X| |O|
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|O|O| |    // Player 2 makes a move at (1, 0).
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|X| |X|
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​
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toe.move(2, 1, 1); -> Returns 1 (player 1 wins)
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|X| |O|
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|O|O| |    // Player 1 makes a move at (2, 1).
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|X|X|X|
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Follow up:
Could you do better than O(n2) permove()operation?
2. Implementation
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class TicTacToe {
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    int[] rows;
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    int[] cols;
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    int diagnol;
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    int antiDiagnol;
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    int n;
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​
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    /** Initialize your data structure here. */
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    public TicTacToe(int n) {
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        rows = new int[n];
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        cols = new int[n];
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        this.n = n;
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    }
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​
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    /** Player {player} makes a move at ({row}, {col}).
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        @param row The row of the board.
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        @param col The column of the board.
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        @param player The player, can be either 1 or 2.
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        @return The current winning condition, can be either:
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                0: No one wins.
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                1: Player 1 wins.
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                2: Player 2 wins. */
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    public int move(int row, int col, int player) {
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        int toAdd = player == 1 ? 1 : -1;
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​
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        rows[row] += toAdd;
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        cols[col] += toAdd;
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        if (row == col) {
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            diagnol += toAdd;
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        }
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​
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        if (col == n - row - 1) {
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            antiDiagnol += toAdd;
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        }
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​
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        if (Math.abs(rows[row]) == n || Math.abs(cols[col]) == n
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            || Math.abs(diagnol) == n || Math.abs(antiDiagnol) == n) {
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            return player;
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        }
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        return 0;
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    }
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}
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​
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/**
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 * Your TicTacToe object will be instantiated and called as such:
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 * TicTacToe obj = new TicTacToe(n);
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 * int param_1 = obj.move(row,col,player);
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 */
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3. Time & Space Complexity
move(): 时间复杂度O(1), 空间复杂度O(n)

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Contents

348. Design Tic-Tac-Toe

1. Question

2. Implementation

3. Time & Space Complexity

348. Design Tic-Tac-Toe​

1. Question

2. Implementation

3. Time & Space Complexity

348. Design Tic-Tac-Toe