标签:abs amp 估计 dia struct animate grid block nbsp
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:
move()
operation can be done in O(1)?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 */
标签:abs amp 估计 dia struct animate grid block nbsp
原文地址:http://www.cnblogs.com/EdwardLiu/p/6182399.html