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

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

Design Tic-Tac Toe的更多相关文章

  1. Principle of Computing (Python)学习笔记(7) DFS Search + Tic Tac Toe use MiniMax Stratedy

    1. Trees Tree is a recursive structure. 1.1 math nodes https://class.coursera.org/principlescomputin ...

  2. POJ 2361 Tic Tac Toe

    题目:给定一个3*3的矩阵,是一个井字过三关游戏.开始为X先走,问你这个是不是一个合法的游戏.也就是,现在这种情况,能不能出现.如果有人赢了,那应该立即停止.那么可以知道X的步数和O的步数应该满足x= ...

  3. 【leetcode】1275. Find Winner on a Tic Tac Toe Game

    题目如下: Tic-tac-toe is played by two players A and B on a 3 x 3 grid. Here are the rules of Tic-Tac-To ...

  4. 2019 GDUT Rating Contest III : Problem C. Team Tic Tac Toe

    题面: C. Team Tic Tac Toe Input file: standard input Output file: standard output Time limit: 1 second M ...

  5. [CareerCup] 17.2 Tic Tac Toe 井字棋游戏

    17.2 Design an algorithm to figure out if someone has won a game oftic-tac-toe. 这道题让我们判断玩家是否能赢井字棋游戏, ...

  6. Epic - Tic Tac Toe

    N*N matrix is given with input red or black.You can move horizontally, vertically or diagonally. If ...

  7. python 井字棋(Tic Tac Toe)

    说明 用python实现了井字棋,整个框架是本人自己构思的,自认为比较满意.另外,90%+的代码也是本人逐字逐句敲的. minimax算法还没完全理解,所以参考了这里的代码,并作了修改. 特点 可以选 ...

  8. ACM-Team Tic Tac Toe

    我的代码: #include <bits/stdc++.h> using namespace std; int main() { char a[3][3]; int i,j=0; for( ...

  9. LeetCode 5275. 找出井字棋的获胜者 Find Winner on a Tic Tac Toe Game

    地址 https://www.acwing.com/solution/LeetCode/content/6670/ 题目描述A 和 B 在一个 3 x 3 的网格上玩井字棋. 井字棋游戏的规则如下: ...

  10. [LeetCode] Design Tic-Tac-Toe 设计井字棋游戏

    Design a Tic-tac-toe game that is played between two players on a n x n grid. You may assume the fol ...

随机推荐

  1. 廖雪峰js教程笔记12 用DOM更新 innerHMTL 和修改css样式

    拿到一个DOM节点后,我们可以对它进行更新. 可以直接修改节点的文本,方法有两种: 一种是修改innerHTML属性,这个方式非常强大,不但可以修改一个DOM节点的文本内容,还可以直接通过HTML片段 ...

  2. Node.js-部署【1】-防火墙端口的配置

    原来以为,Node.js部署以后,要手动配置防火墙端口,结果不需要,外网可以访问,看来是自动配好了,真是考虑周到,给我一个大大的惊喜.

  3. 冰球项目日志2-yjw

    我们小组在12.31号进行了讨论,确定了基本的任务及分工,后面是元旦放假,大家没有做很多的东西,我也是把自己分工的部分方案想了下. 后面在1.3号,我们会再进行一次小组讨论,确定下最终的方案,然后进行 ...

  4. Mac下抓包

    Wireshark针对UNIX Like系统的GUI发行版界面采用的是X Window(1987年更改X版本到X11).Mac OS X在Mountain Lion之后放弃X11,取而代之的是开源的X ...

  5. 解决Winform应用程序中窗体背景闪烁的问题

    本文转载:https://my.oschina.net/Tsybius2014/blog/659742 我的操作系统是Win7,使用的VS版本是VS2012,文中的代码都是C#代码. 这几天遇到一个问 ...

  6. hud 5876 2016 ACM/ICPC Asia Regional Dalian Online

    题意:给一个图 给定一个点s 求补图中s点到达各个点的最短路 思路:从s点开始bfs 在图中与s点有连接的都是在补图中不能直接到达的点 反之在补图中都是可以直接到达的点 由此bfs ((( 诡异的写法 ...

  7. vs2015帮助文档

    1)注释快捷键: CTRL + K - CTRL + C (注释) CTRL + K 然后 CTRL + U (取消注释) shift+"*"---------整段(取消)注释 2 ...

  8. synchronized的实现原理和应用

    在多线程并发编程中synchronized是元老级的角色,人多称重量级锁. synchronized实现同步的基础:Java中的每一个对象都可以作为锁.具体表现有如下3种: 1.对于普通同步方法,锁时 ...

  9. java异常捕获

    类ExampleA继承Exception,类ExampleB继承ExampleA. 有如下代码片断: try { throw new ExampleB("b") } catch(E ...

  10. iOS相关笔记

    #协议[1] [2] @property (nonatomic, assign) id<EveryFrameDelegate> delegate; 表明,这个delegate是一个需要实现 ...