题目：
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() 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.解答：
一开始其实就想到了hint, 作出了下面的解法：public class TicTacToe { int[][] grid; int n; /** Initialize your data structure here. */ public TicTacToe(int n) { grid = new int[n][n]; this.n = n; } public boolean check(int row, int col, int len) { boolean hori = true, verti = true, diag1 = true, diag2 = true; //check horizontal for (int i = 0; i < len - 1; i++) { if (grid[row][i] != grid[row][i + 1]) { hori = false; } } //check vertical for (int j = 0; j < len - 1; j++) { if (grid[j][col] != grid[j + 1][col]) { verti = false; } } //check diagonals if (row == col) { for (int i = 0; i < len - 1; i++) { if (grid[i][i] != grid[i + 1][i + 1]) { diag1 = false; } } } else { diag1 = false; } if (row + col == len - 1) { for (int i = 0; i < len - 1; i++) { if (grid[i][len - 1 - i] != grid[i + 1][len - 2 - i]) { diag2 = false; } } } else { diag2 = false; } return hori || verti || diag1 || diag2; } /** 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) { grid[row][col] = player; if (check(row, col, n)) return player; return 0; }} 这个解法冗余在check行个列的时候，每一次都要再扫一遍这一行这一列，所以如果只有两个player，可以把这两个player记作1, －1。当有一行完全只有这两个player中的其中一个人时，sum的绝对值应该等于这个数列的长度，这样就不需要每次再扫一遍数组。代码如下：
public class TicTacToe { int[] rows, cols; int diagonal, antiDiagonal; int len; /** Initialize your data structure here. */ public TicTacToe(int n) { rows = new int[n]; cols = new int[n]; this.len = 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) { //Take player 1 and 2 as value of 1 and -1; //Every time we only do adding, dont need to re-scan the whole line int toAdd = player == 1 ? 1 : -1; rows[row] += toAdd; cols[col] += toAdd; if (row == col) diagonal += toAdd; if (row == len - 1 - col) antiDiagonal += toAdd; if (Math.abs(rows[row]) == len || Math.abs(cols[col]) == len || Math.abs(diagonal) == len || Math.abs(antiDiagonal) == len) { return player; } return 0; }}