[LeetCode] 348. Design Tic-Tac-Toe 设计井字棋游戏
Design a Tic-tac-toe game that is played between two players on a n x n grid.
You may assume the following rules:
- 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. 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()
Hint:
Could you trade extra space such that move() operation can be done in O(1)?
You need two arrays: int rows[n], int cols[n], plus two variables: diagonal, anti_diagonal.
CareerCup上的原题,请参见我之前的博客17.2 Tic Tac Toe。我们首先来O(n2)的解法,这种方法的思路很straightforward,就是建立一个nxn大小的board,其中0表示该位置没有棋子,1表示玩家1放的子,2表示玩家2。那么棋盘上每增加一个子,我们都扫描当前行列,对角线,和逆对角线,看看是否有三子相连的情况,有的话则返回对应的玩家,没有则返回0,参见代码如下:
解法一:
class TicTacToe { public: /** Initialize your data structure here. */ TicTacToe(int n) { board.resize(n, vector<int>(n, 0)); } int move(int row, int col, int player) { board[row][col] = player; int i = 0, j = 0, n = board.size(); for (j = 1; j < n; ++j) { if (board[row][j] != board[row][j - 1]) break; } if (j == n) return player; for (i = 1; i < n; ++i) { if (board[i][col] != board[i - 1][col]) break; } if (i == n) return player; if (row == col) { for (i = 1; i < n; ++i) { if (board[i][i] != board[i - 1][i - 1]) break; } if (i == n) return player; } if (row + col == n - 1) { for (i = 1; i < n; ++i) { if (board[n - i - 1][i] != board[n - i][i - 1]) break; } if (i == n) return player; } return 0; } private: vector<vector<int>> board; };
Follow up中让我们用更高效的方法,那么根据提示中的,我们建立一个大小为n的一维数组rows和cols,还有变量对角线diag和逆对角线rev_diag,这种方法的思路是,如果玩家1在第一行某一列放了一个子,那么rows[0]自增1,如果玩家2在第一行某一列放了一个子,则rows[0]自减1,那么只有当rows[0]等于n或者-n的时候,表示第一行的子都是一个玩家放的,则游戏结束返回该玩家即可,其他各行各列,对角线和逆对角线都是这种思路,参见代码如下:
解法二:
class TicTacToe { public: /** Initialize your data structure here. */ TicTacToe(int n): rows(n), cols(n), N(n), diag(0), rev_diag(0) {} int move(int row, int col, int player) { int add = player == 1 ? 1 : -1; rows[row] += add; cols[col] += add; diag += (row == col ? add : 0); rev_diag += (row == N - col - 1 ? add : 0); return (abs(rows[row]) == N || abs(cols[col]) == N || abs(diag) == N || abs(rev_diag) == N) ? player : 0; } private: vector<int> rows, cols; int diag, rev_diag, N; };
参考资料:
https://leetcode.com/problems/design-tic-tac-toe/
https://discuss.leetcode.com/topic/44548/java-o-1-solution-easy-to-understand
https://discuss.leetcode.com/topic/44605/c-time-o-1-space-o-n-short-simple-solution