原文链接: https://leetcode-cn.com/problems/tic-tac-toe-lcci

英文原文

Design an algorithm to figure out if someone has won a game of tic-tac-toe. Input is a string array of size N x N, including characters " ", "X" and "O", where " " represents a empty grid.

The rules of tic-tac-toe are as follows:

• Players place characters into an empty grid(" ") in turn.
• The first player always place character "O", and the second one place "X".
• Players are only allowed to place characters in empty grid. Replacing a character is not allowed.
• If there is any row, column or diagonal filled with N same characters, the game ends. The player who place the last charater wins.
• When there is no empty grid, the game ends.
• If the game ends, players cannot place any character further.

If there is any winner, return the character that the winner used. If there's a draw, return "Draw". If the game doesn't end and there is no winner, return "Pending".

Example 1:

Input:  board = ["O X"," XO","X O"]
Output:  "X"

Example 2:

Input:  board = ["OOX","XXO","OXO"]
Output:  "Draw"
Explanation:  no player wins and no empty grid left

Example 3:

Input:  board = ["OOX","XXO","OX "]
Output:  "Pending"
Explanation:  no player wins but there is still a empty grid

Note:

• 1 <= board.length == board[i].length <= 100
• Input follows the rules.

中文题目

设计一个算法，判断玩家是否赢了井字游戏。输入是一个 N x N 的数组棋盘，由字符" "，"X"和"O"组成，其中字符" "代表一个空位。

以下是井字游戏的规则：

• 玩家轮流将字符放入空位（" "）中。
• 第一个玩家总是放字符"O"，且第二个玩家总是放字符"X"。
• "X"和"O"只允许放置在空位中，不允许对已放有字符的位置进行填充。
• 当有N个相同（且非空）的字符填充任何行、列或对角线时，游戏结束，对应该字符的玩家获胜。
• 当所有位置非空时，也算为游戏结束。
• 如果游戏结束，玩家不允许再放置字符。

如果游戏存在获胜者，就返回该游戏的获胜者使用的字符（"X"或"O"）；如果游戏以平局结束，则返回 "Draw"；如果仍会有行动（游戏未结束），则返回 "Pending"。

示例 1：

输入： board = ["O X"," XO","X O"]
输出： "X"

示例 2：

输入： board = ["OOX","XXO","OXO"]
输出： "Draw"
解释： 没有玩家获胜且不存在空位

示例 3：

输入： board = ["OOX","XXO","OX "]
输出： "Pending"
解释： 没有玩家获胜且仍存在空位

提示：

• 1 <= board.length == board[i].length <= 100
• 输入一定遵循井字棋规则

通过代码

高赞题解

// 别人的想法，我加个注释，希望大家看得清楚一点

class Solution {
    public String tictactoe(String[] board) {

        int length = board.length;
        int heng = 0; //横的和
        int zong = 0; //纵的和
        int left = 0; //左斜线
        int right = 0; //右斜线
        boolean flag = false; //记录有没有空格

        for (int i = 0; i < length; i++) {

            heng = 0; zong = 0;

            for (int j = 0; j < length; j++) {

                heng = heng +  (int) board[i].charAt(j);
                zong = zong + (int) board[j].charAt(i);

                if(board[i].charAt(j) == ' ') flag = true;

            }

            //横纵检查
            if (heng == (int)'X' * length || zong == (int)'X' * length) return "X";
            if (heng == (int)'O' * length || zong == (int)'O' * length) return "O";

            //两条斜线上的相加
            left = left + (int)board[i].charAt(i);
            right = right + (int)board[i].charAt(length - i - 1);

        }

        //两条斜线检查
        if (left == (int)'X' * length || right == (int)'X' * length) return "X";
        if (left == (int)'O' * length || right == (int)'O' * length) return "O";

        if (flag) return "Pending";
        return "Draw";

    }
}

统计信息

通过次数	提交次数	AC比率
7633	16601	46.0%

提交历史

提交时间	提交结果	执行时间	内存消耗	语言