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:
- 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.
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
- 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 ...
- 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= ...
- 【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 ...
- 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 ...
- [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, ...
- Epic - Tic Tac Toe
N*N matrix is given with input red or black.You can move horizontally, vertically or diagonally. If ...
- 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 ...
- 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( ...
- 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: ...
- [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
- 2015ACM/ICPC亚洲区沈阳站
5510 Bazinga 题意:给出n个字符串,求满足条件的最大下标值或层数 条件:该字符串之前存在不是 它的子串 的字符串 求解si是不是sj的子串,可以用kmp算法之类的. strstr是黑科技, ...
- 后台动态生成GridView列和模版
考虑到很多数据源是不确定的,所以这时无法在前台设置gridview的表头,需要在后台动态指定并绑定数据. 前台代码如下: <%@ Page Title="主页" Langua ...
- Linux下定时任务配置-crontab
实际中经常有一些任务需要定期执行,人工操作比较麻烦,如果定时执行将会省去很多人力,还可以在一些资源占用不多的时间段执行,linux下crontab命令就实现了这一便捷的功能,实现脚本的自动化运行. 常 ...
- Leetcode 4Sum
Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c + d = tar ...
- 尝试封装适用于权限管理的通用API
谈谈我对权限系统的简单理解 最近一段时间在研究权限系统,在园子里看到个很牛逼的开源的基于DDD-Lite的权限管理系统,并有幸加入了作者的QQ群,呵呵,受到了很大的影响.对于权限管理我有我自己的一些简 ...
- python序列
序列基础 序列:python包含6种内建的序列,常用的有:列表.元组.字符串.列表可以修改,元组和字符串不能修改. 索引:从0开始递增,通过索引获取元素:可使用负数索引,从右至左.最后1个元素的位置编 ...
- SQLite部署-无法加载 DLL“SQLite.Interop.dll”: 找不到指定的模块
近期刚使用SQLite,主要引用的是System.Data.SQLite.dll这个dll,在部署到测试环境时报无法加载 DLL“SQLite.Interop.dll”: 找不到指定的模块. (异常来 ...
- jQuery实现全选效果【转】
这是一段用jquery实现全选的代码,主要思路如下: 1.所有的复选框都有单击事件,所有效果都是在单击事件下实现的 2.全选复选框所实现的功能与其他复选选项实现的功能不同,所有在单击事件内做一个判断, ...
- Winform TextBox中只能输入数字的几种常用方法(C#)
方法一: private void tBox_KeyPress(object sender, KeyPressEventArgs e) { ; //禁止空格键 )) return; //处理负数 if ...
- Android四大组件--ContentProvider详解(转)
一.相关ContentProvider概念解析: 1.ContentProvider简介在Android官方指出的Android的数据存储方式总共有五种,分别是:Shared Preferences. ...