[Leetcode] 348. Design Tic-Tac-Toe 解题报告

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:

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:

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?

思路：

这道题目的思路没有什么特别奇怪的。只是我开始用了二维数组来表示棋盘，这样空间复杂度就成为O(n^2)，move函数的时间复杂度是O(n)。后来发现一种更高效的实现方法：用两个数组和两个整数表示一个player目前达到的状态。这样空间复杂度就可以降低到O(n)，而move的时间复杂度竟然可以降低到O(1)。

这种面试题到处都是坑啊！最优算法往往未必就能一下子想到。

代码：

class TicTacToe {public:    /** Initialize your data structure here. */    TicTacToe(int n) {        rows_first.resize(n, 0);        rows_second.resize(n, 0);        cols_first.resize(n, 0);        cols_second.resize(n, 0);        diag_first = diag_second = anti_diag_first = anti_diag_second = 0;        size = 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. */    int move(int row, int col, int player) {        if(player == 1) {            if(++rows_first[row] == size)                           return 1;            if(++cols_first[col] == size)                           return 1;            if(row == col && ++diag_first == size)                  return 1;            if(row + col == size - 1 && ++anti_diag_first == size)  return 1;        }        else {            if(++rows_second[row] == size)                          return 2;            if(++cols_second[col] == size)                          return 2;            if(row == col && ++diag_second == size)                 return 2;            if(row + col == size - 1 && ++anti_diag_second == size) return 2;        }        return 0;    }private:    vector<int> rows_first, rows_second;    vector<int> cols_first, cols_second;    int diag_first, diag_second, anti_diag_first, anti_diag_second;    int size;};/** * Your TicTacToe object will be instantiated and called as such: * TicTacToe obj = new TicTacToe(n); * int param_1 = obj.move(row,col,player); */