• LeetCode-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?

    Solution:

     1 public class TicTacToe {
     2     int[] rowCount;
     3     int[] colCount;
     4     int leftDiagCount, rightDiagCount;
     5     int winner;
     6     int num;
     7 
     8     /** Initialize your data structure here. */
     9     public TicTacToe(int n) {
    10         num = n;
    11         rowCount = new int[n];
    12         colCount = new int[n];
    13         leftDiagCount = rightDiagCount = 0;
    14         winner = 0;
    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         if (winner!=0) return winner;
    27         
    28         int delta = (player==1) ? 1 : -1;
    29         rowCount[row] += delta;
    30         colCount[col] += delta;
    31         if (row==col) leftDiagCount += delta;
    32         if (row+col == num-1) rightDiagCount += delta; 
    33         
    34         if (Math.abs(rowCount[row]) == num || Math.abs(colCount[col])==num || Math.abs(leftDiagCount) == num || Math.abs(rightDiagCount) == num){
    35             winner = player;
    36         }
    37         
    38         return winner;
    39     }
    40 }
    41 
    42 /**
    43  * Your TicTacToe object will be instantiated and called as such:
    44  * TicTacToe obj = new TicTacToe(n);
    45  * int param_1 = obj.move(row,col,player);
    46  */
  • 相关阅读:
    [转]跨语言通信方案比较
    C#三种定时器
    Java优化技巧
    websocket初探
    [转]远远走来一个绿茶婊
    赠与今年的大学毕业生-----------胡适
    HDU3068 回文串 Manacher算法
    OpenCV安装与配置
    tkinter事件机制
    哈夫曼压缩
  • 原文地址:https://www.cnblogs.com/lishiblog/p/5760514.html
Copyright © 2020-2023  润新知