Assume the following rules are for the tic-tac-toe game on an n x n board between two players:
- A move is guaranteed to be valid and is placed on an empty block.
- Once a winning condition is reached, no more moves are allowed.
- A player who succeeds in placing
nof their marks in a horizontal, vertical, or diagonal row wins the game.
Implement the TicTacToe class:
-
TicTacToe(int n)Initializes the object the size of the boardn. -
int move(int row, int col, int player)Indicates that player with idplayerplays at the cell(row, col)of the board. The move is guaranteed to be a valid move.
Follow up:
Could you do better than O(n2) per move() operation?
Example 1:
Input ["TicTacToe", "move", "move", "move", "move", "move", "move", "move"] [[3], [0, 0, 1], [0, 2, 2], [2, 2, 1], [1, 1, 2], [2, 0, 1], [1, 0, 2], [2, 1, 1]] Output [null, 0, 0, 0, 0, 0, 0, 1] Explanation TicTacToe ticTacToe = new TicTacToe(3); Assume that player 1 is "X" and player 2 is "O" in the board. ticTacToe.move(0, 0, 1); // return 0 (no one wins) |X| | | | | | | // Player 1 makes a move at (0, 0). | | | | ticTacToe.move(0, 2, 2); // return 0 (no one wins) |X| |O| | | | | // Player 2 makes a move at (0, 2). | | | | ticTacToe.move(2, 2, 1); // return 0 (no one wins) |X| |O| | | | | // Player 1 makes a move at (2, 2). | | |X| ticTacToe.move(1, 1, 2); // return 0 (no one wins) |X| |O| | |O| | // Player 2 makes a move at (1, 1). | | |X| ticTacToe.move(2, 0, 1); // return 0 (no one wins) |X| |O| | |O| | // Player 1 makes a move at (2, 0). |X| |X| ticTacToe.move(1, 0, 2); // return 0 (no one wins) |X| |O| |O|O| | // Player 2 makes a move at (1, 0). |X| |X| ticTacToe.move(2, 1, 1); // return 1 (player 1 wins) |X| |O| |O|O| | // Player 1 makes a move at (2, 1). |X|X|X|
Constraints:
2 <= n <= 100- player is
1or2. 1 <= row, col <= n-
(row, col)are unique for each different call tomove. - At most
n2calls will be made tomove.
设计井字棋。
请在 n × n 的棋盘上，实现一个判定井字棋（Tic-Tac-Toe）胜负的神器，判断每一次玩家落子后，是否有胜出的玩家。
在这个井字棋游戏中，会有 2 名玩家，他们将轮流在棋盘上放置自己的棋子。
在实现这个判定器的过程中，你可以假设以下这些规则一定成立：
1. 每一步棋都是在棋盘内的，并且只能被放置在一个空的格子里；
2. 一旦游戏中有一名玩家胜出的话，游戏将不能再继续；
3. 一个玩家如果在同一行、同一列或者同一斜对角线上都放置了自己的棋子，那么他便获得胜利。
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/design-tic-tac-toe
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
这是一道设计题。题目的 follow up 问能不能对于 move() 函数，给出优于 O(n^2) 的思路，我这里给出一个最优解。这个游戏给的棋盘是一个长度为 N 的正方形，只有两个玩家，如果两个玩家有任何一个玩家的棋子摆满了某一行，某一列或者某一条对角线，则判定为胜利，那么我们可以给两个玩家分别赋值 1 和 -1，来判断到底是谁走棋。如果某一行/某一列被同一个玩家占满，则那一行/那一列的sum会是 N 或者 -N。对于对角线也是一样，只是这里对角线的计算有些复杂，但是注意因为是正方形所以计算相对简单
- 正对角线diagonal - 只要判断是否 row == col 即可
- 反对角线antiDiagonal - 判断是否 col == N - row - 1即可
时间O(1)
空间O(n^2)
Java实现
1 class TicTacToe { 2 private int[] rows; 3 private int[] cols; 4 private int diagonal; 5 private int antiDiagonal; 6 7 /** Initialize your data structure here. */ 8 public TicTacToe(int n) { 9 rows = new int[n]; 10 cols = new int[n]; 11 } 12 13 /** Player {player} makes a move at ({row}, {col}). 14 @param row The row of the board. 15 @param col The column of the board. 16 @param player The player, can be either 1 or 2. 17 @return The current winning condition, can be either: 18 0: No one wins. 19 1: Player 1 wins. 20 2: Player 2 wins. */ 21 public int move(int row, int col, int player) { 22 int toAdd = player == 1 ? 1 : -1; 23 rows[row] += toAdd; 24 cols[col] += toAdd; 25 if (row == col) { 26 diagonal += toAdd; 27 } 28 if (col == (cols.length - row - 1)) { 29 antiDiagonal += toAdd; 30 } 31 32 int size = rows.length; 33 if (Math.abs(rows[row]) == size || Math.abs(cols[col]) == size || Math.abs(diagonal) == size 34 || Math.abs(antiDiagonal) == size) { 35 return player; 36 } 37 return 0; 38 } 39 } 40 41 /** 42 * Your TicTacToe object will be instantiated and called as such: 43 * TicTacToe obj = new TicTacToe(n); 44 * int param_1 = obj.move(row,col,player); 45 */
