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  */

 

posted @ 2016-08-11 12:36  LiBlog  阅读(233)  评论(0)  编辑  收藏  举报