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348.Design-Tic-Tac-Toe

348. Design Tic Tac Toe

题目地址

题目描述

Design a Tic-tac-toe game that is played between two players on a n x n grid.
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You may assume the following rules:
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A move is guaranteed to be valid and is placed on an empty block.
Once a winning condition is reached, no more moves is allowed.
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.
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TicTacToe toe = new TicTacToe(3);
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toe.move(0, 0, 1); -> Returns 0 (no one wins)
|X| | |
| | | | // Player 1 makes a move at (0, 0).
| | | |
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toe.move(0, 2, 2); -> Returns 0 (no one wins)
|X| |O|
| | | | // Player 2 makes a move at (0, 2).
| | | |
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toe.move(2, 2, 1); -> Returns 0 (no one wins)
|X| |O|
| | | | // Player 1 makes a move at (2, 2).
| | |X|
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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?

代码

Approach #1

class TicTacToe {
private int[] rows;
private int[] cols;
private int diagonal;
private int antiDiagonal;
​
public TicTacToe(int n) {
rows = new int[n];
cols = new int[n];
}
​
public int move(int row, int col, int player) {
int size = rows.length;
int toAdd = (player == 1 ? 1 : -1);
int target = (player == 1 ? size : -size);
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rows[row] += toAdd;
cols[col] += toAdd;
if (row == col) {
diagonal += toAdd;
}
​
if (col + row == size - 1) {
antiDiagonal += toAdd;
}
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if (rows[row] == target ||
cols[col] == target ||
diagonal == target ||
antiDiagonal == target) {
​
return player;
}
​
return 0;
}
}