zoukankan      html  css  js  c++  java
  • 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:
    
    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:

    1. Could you trade extra space such that move() operation can be done in O(1)?
    2. You need two arrays: int rows[n], int cols[n], plus two variables: diagonal, anti_diagonal.

    看了hint之后想到:既然用数组的话,一行一列都是一个element来代表,估计这个element是要用sum了,那么,能不能用sum来代表一行,使它有且只有一种可能,全部是player1完成的/全部是Player2;所以想到了是Player1就+1,是player2就-1,看最后sum是不是n,或者-n;n的情况只有一种情况这一行全是player1。因为说了不会有invalid move, 所以情况是唯一的

     1 public class TicTacToe {
     2     int[] rows;
     3     int[] cols;
     4     int diagonal;
     5     int anti_diagonal;
     6     int size;
     7 
     8     /** Initialize your data structure here. */
     9     public TicTacToe(int n) {
    10         rows = new int[n];
    11         cols = new int[n];
    12         diagonal = 0;
    13         anti_diagonal = 0;
    14         size = n;
    15     }
    16     
    17     /** Player {player} makes a move at ({row}, {col}).
    18         @param row The row of the board.
    19         @param col The column of the board.
    20         @param player The player, can be either 1 or 2.
    21         @return The current winning condition, can be either:
    22                 0: No one wins.
    23                 1: Player 1 wins.
    24                 2: Player 2 wins. */
    25     public int move(int row, int col, int player) {
    26         int change = (player==1? 1 : -1);
    27         rows[row] += change;
    28         cols[col] += change;
    29         if (row == col) diagonal += change;
    30         if (row + col == size-1) anti_diagonal += change;
    31         if (Math.abs(rows[row])==size || Math.abs(cols[col])==size || Math.abs(diagonal)==size || Math.abs(anti_diagonal)==size)
    32             return player;
    33         return 0;
    34     }
    35 }
    36 
    37 /**
    38  * Your TicTacToe object will be instantiated and called as such:
    39  * TicTacToe obj = new TicTacToe(n);
    40  * int param_1 = obj.move(row,col,player);
    41  */
  • 相关阅读:
    vim基本配置
    docker-compose.yml的使用
    centos7安装docker-compose
    gitlab常用命令
    搭建gitlab服务
    jdk安装
    安装部署mongodb
    源码安装部署redis
    febric的使用
    IntelliJ IDEA 快捷键终极大全,速度收藏!
  • 原文地址:https://www.cnblogs.com/EdwardLiu/p/6182399.html
Copyright © 2011-2022 走看看