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(n^2) per move() operation?
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.
解法1: 暴力解法,每走一步,对所走点的水平,竖直,对角线,反对角线进行检查是否满足条件。
解法2: 根据提示,分别建立水平,竖直两个数组,以及对角线,反对角线两个变量。每走一步分别对这几个进行判断,一个玩家加1,一个玩家-1,如果下棋子的点的水平或者垂直数组里的元素的值等于n, 或者对角线的值的绝对值等于n,那么就返回此时下棋子的选手赢。
Java:
public class TicTacToe { int[][] matrix; /** Initialize your data structure here. */ public TicTacToe(int n) { matrix = new int[n][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) { matrix[row][col]=player; //check row boolean win=true; for(int i=0; i<matrix.length; i++){ if(matrix[row][i]!=player){ win=false; break; } } if(win) return player; //check column win=true; for(int i=0; i<matrix.length; i++){ if(matrix[i][col]!=player){ win=false; break; } } if(win) return player; //check back diagonal win=true; for(int i=0; i<matrix.length; i++){ if(matrix[i][i]!=player){ win=false; break; } } if(win) return player; //check forward diagonal win=true; for(int i=0; i<matrix.length; i++){ if(matrix[i][matrix.length-i-1]!=player){ win=false; break; } } if(win) return player; return 0; } }
Java:
public class TicTacToe { int[] rows; int[] cols; int dc1; int dc2; int n; /** Initialize your data structure here. */ public TicTacToe(int n) { this.n=n; this.rows=new int[n]; this.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 val = (player==1?1:-1); rows[row]+=val; cols[col]+=val; if(row==col){ dc1+=val; } if(col==n-row-1){ dc2+=val; } if(Math.abs(rows[row])==n || Math.abs(cols[col])==n || Math.abs(dc1)==n || Math.abs(dc2)==n){ return player; } return 0; } }
Python:
class TicTacToe(object): def __init__(self, n): """ Initialize your data structure here. :type n: int """ self.__size = n self.__rows = [[0, 0] for _ in xrange(n)] self.__cols = [[0, 0] for _ in xrange(n)] self.__diagonal = [0, 0] self.__anti_diagonal = [0, 0] def move(self, row, col, player): """ 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. :type row: int :type col: int :type player: int :rtype: int """ i = player - 1 self.__rows[row][i] += 1 self.__cols[col][i] += 1 if row == col: self.__diagonal[i] += 1 if col == len(self.__rows) - row - 1: self.__anti_diagonal[i] += 1 if any(self.__rows[row][i] == self.__size, self.__cols[col][i] == self.__size, self.__diagonal[i] == self.__size, self.__anti_diagonal[i] == self.__size): return player return 0
C++:
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; };
C++:
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; };
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