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

题目链接: https://leetcode.com/problems/design-tic-tac-toe/

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: 搞不懂为啥他说在O(n*n)以内解决, 感觉不可能到O(n*n)的复杂度啊, 每一步先判断横竖两行, 如果能够排满就赢了, 否则如果这个位置在两个对角线上再判断对角线是否排满即可. 时间复杂度O(n) or O(1).

代码如下:

class TicTacToe {public:    /** Initialize your data structure here. */    TicTacToe(int n):mp(vector<vector<int>>(n, vector<int>(n, 0))), 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. */    int move(int row, int col, int player) {        mp[row][col] = player;        int i;        for(i=0; i<N; i++) if(mp[row][i]!=player) break;        if(i==N) return player;        for(i=0; i<N; i++) if(mp[i][col]!=player) break;        if(i==N) return player;        if(row==col)            for(i=0; i<N; i++) if(mp[i][i]!=player) break;        if(i==N) return player;        if(row+col==N-1)            for(i=0; i<N; i++) if(mp[N-i-1][i]!=player) break;        if(i==N) return player;        return false;    }private:    vector<vector<int>> mp;    int N;};/** * Your TicTacToe object will be instantiated and called as such: * TicTacToe obj = new TicTacToe(n); * int param_1 = obj.move(row,col,player); */

class TicTacToe {public:    /** Initialize your data structure here. */    TicTacToe(int n):rows(vector<int>(n, 0)), cols(vector<int>(n, 0)) {            }        /** 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) {        int flag = (player==1?1:-1), n = rows.size();        rows[row] += flag, cols[col]+=flag;        if(row==col) antiDiagnal += flag;        if(row+col+1 == n) diagnal += flag;        if(abs(rows[row])==n || abs(cols[col])==n           ||abs(diagnal)==n || abs(antiDiagnal)==n)             return player;        return 0;    }private:    vector<int> rows, cols;    int diagnal=0, antiDiagnal=0;};/** * Your TicTacToe object will be instantiated and called as such: * TicTacToe obj = new TicTacToe(n); * int param_1 = obj.move(row,col,player); */