資訊專欄INFORMATION COLUMN
Problem
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?
class TicTacToe {
private int[] rows;
private int[] cols;
private int diagonal;
private int antiDiagonal;
private int size;
/** Initialize your data structure here. */
public TicTacToe(int n) {
rows = new int[n];
cols = new int[n];
diagonal = 0;
antiDiagonal = 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 num = player == 1 ? 1 : -1;
rows[row] += num;
cols[col] += num;
if (row == col) diagonal += num;
if (row == size-1-col) antiDiagonal += num;
if (Math.abs(rows[row]) == size ||
Math.abs(cols[col]) == size ||
Math.abs(diagonal) == size ||
Math.abs(antiDiagonal) == size) return player;
return 0;
}
}
文章版權(quán)歸作者所有，未經(jīng)允許請勿轉(zhuǎn)載,若此文章存在違規(guī)行為，您可以聯(lián)系管理員刪除。
轉(zhuǎn)載請注明本文地址：http://m.hztianpu.com/yun/72457.html
摘要：題目鏈接這道題找是否有贏的方法和相似，稍微簡化了。統(tǒng)計行列和兩個對角線的情況，兩個分別用和來記。然后判斷是否有一個人贏只需要的復(fù)雜度。當(dāng)然這么做的前提是假設(shè)所有的都是的，棋盤一個地方已經(jīng)被占用了，就不能走那個地方了。 348. Design Tic-Tac-Toe 題目鏈接：https://leetcode.com/problems... 這道題找是否有player贏的方法和N-Que...
摘要：當(dāng)有一行完全只有這兩個中的其中一個人時，的絕對值應(yīng)該等于這個數(shù)列的長度，這樣就不需要每次再掃一遍數(shù)組。 題目：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 b...
摘要：我們在前文中考慮的那張圖就來自這篇文章，之后我們會用剪枝算法來改進(jìn)之前的解決方案。剪枝算法的實(shí)現(xiàn)接下來討論如何修改前面實(shí)現(xiàn)的算法，使其變?yōu)榧糁λ惴ā，F(xiàn)在我們已經(jīng)有了現(xiàn)成的和剪枝算法，只要加上一點(diǎn)兒細(xì)節(jié)就能完成這個游戲了。 前段時間用 React 寫了個2048 游戲來練練手，準(zhǔn)備用來回顧下 React 相關(guān)的各種技術(shù)，以及試驗一下新技術(shù)。在寫這個2048的過程中，我考慮是否可以在其中加...
摘要：我們在前文中考慮的那張圖就來自這篇文章，之后我們會用剪枝算法來改進(jìn)之前的解決方案。剪枝算法的實(shí)現(xiàn)接下來討論如何修改前面實(shí)現(xiàn)的算法，使其變?yōu)榧糁λ惴ā，F(xiàn)在我們已經(jīng)有了現(xiàn)成的和剪枝算法，只要加上一點(diǎn)兒細(xì)節(jié)就能完成這個游戲了。 前段時間用 React 寫了個2048 游戲來練練手，準(zhǔn)備用來回顧下 React 相關(guān)的各種技術(shù)，以及試驗一下新技術(shù)。在寫這個2048的過程中，我考慮是否可以在其中加...
摘要：題目鏈接題目分析設(shè)計一個哈希類。需要有添加元素函數(shù)，判斷元素存在的函數(shù)，移除元素函數(shù)。思路這真的沒什么好說的了我把要存的值作為數(shù)組的鍵存儲。最終代碼若覺得本文章對你有用，歡迎用愛發(fā)電資助。 D87 705. Design HashSet 題目鏈接 705. Design HashSet 題目分析 設(shè)計一個哈希類。 需要有add添加元素函數(shù)，contains判斷元素存在的函數(shù)，remov...
閱讀 1654·2019-08-30 13:18
閱讀 1637·2019-08-29 12:19
閱讀 2188·2019-08-26 13:57
閱讀 4211·2019-08-26 13:22
閱讀 1265·2019-08-26 10:35
閱讀 3051·2019-08-23 18:09
閱讀 2597·2019-08-23 17:19
閱讀 746·2019-08-23 17:18
