---

CheckiO 是面向初學者和高級程序員的編碼游戲，使用 Python 和 JavaScript 解決棘手的挑戰和有趣的任務，從而提高你的編碼技能，本博客主要記錄自己用 Python 在闖關時的做題思路和實現代碼，同時也學習學習其他大神寫的代碼。

CheckiO 官網：https://checkio.org/

我的 CheckiO 主頁：https://py.checkio.org/user/TRHX/

CheckiO 題解系列專欄：https://itrhx.blog.csdn.net/category_9536424.html

CheckiO 所有題解源代碼：https://github.com/TRHX/Python-CheckiO-Exercise

---

題目描述

【Xs and Os Referee】：井字游戲，兩個玩家（X和O）輪流在 3×3 的網格上落棋，最先在任意一條直線（水平線，垂直線或對角線）上成功連接三個網格的一方獲勝。在本題中，你將是這個游戲的裁判。你必須判斷游戲是平局還是有人勝出，以及誰將會成為最后的贏家。如果 X 玩家獲勝，返回“X”。如果 O 玩家獲勝，返回“O”。如果比賽是平局，返回“D”。

【鏈接】：https://py.checkio.org/mission/x-o-referee/

【輸入】：“X”，“O”在棋盤上的位置，“.”表示空格（列表）

【輸出】：獲勝的一方，“X”，“O”或“D”（字符串）

【范例】：

checkio(["X.O","XX.","XOO"]) == "X" checkio(["OO.","XOX","XOX"]) == "O" checkio(["OOX","XXO","OXX"]) == "D"

解題思路

判斷誰贏有 6 種情況，橫著 3 種，豎著 3 種，斜著 2 種，我用了最笨的辦法，直接將每顆棋子轉換成列表，依次判斷是否相等就行了，就是不斷的用 if 語句，另外也可以分橫豎和斜著兩種情況來寫代碼，這樣的話，橫豎可以設置一個變量，循環 3 次，利用類似于 game_result[i][0]、game_result[i][1] 來依次比較，需要特別注意的是，要排除三個空格，也就是三個相同 . 的情況

代碼實現

from typing import Listdef checkio(game_result: List[str]) -> str:list2 = []for i in game_result:list2.extend(list(i))if list2[0] != '.' and (list2[0] == list2[3] == list2[6] or list2[0] == list2[4] == list2[8] or list2[0] == list2[1] == list2[2]):return list2[0]elif list2[1] != '.' and (list2[1] == list2[4] == list2[7]):return list2[1]elif list2[2] != '.' and (list2[2] == list2[4] == list2[6] or list2[2] == list2[5] == list2[8]):return list2[2]elif list2[3] != '.' and (list2[3] == list2[4] == list2[5]):return list2[3]elif list2[6] != '.' and (list2[6] == list2[7] == list2[8]):return list2[6]else:return 'D'if __name__ == '__main__':print("Example:")print(checkio(["X.O","XX.","XOO"]))# These "asserts" using only for self-checking and not necessary for auto-testingassert checkio(["X.O","XX.","XOO"]) == "X", "Xs wins"assert checkio(["OO.","XOX","XOX"]) == "O", "Os wins"assert checkio(["OOX","XXO","OXX"]) == "D", "Draw"assert checkio(["O.X","XX.","XOO"]) == "X", "Xs wins again"print("Coding complete? Click 'Check' to review your tests and earn cool rewards!")

大神解答

大神解答 NO.1

def checkio(result):rows = resultcols = map(''.join, zip(*rows))diags = map(''.join, zip(*[(r[i], r[2 - i]) for i, r in enumerate(rows)]))lines = rows + list(cols) + list(diags)return 'X' if ('XXX' in lines) else 'O' if ('OOO' in lines) else 'D'

大神解答 NO.2

# From Daniel Dou with love...def checkio(board):# First we put everything together into a single stringx = "".join(board)# Next we outline the 8 possible winning combinations. combos = ["012", "345", "678", "036", "147", "258", "048", "246"]# We go through all the winning combos 1 by 1 to see if there are any# all Xs or all Os in the combosfor i in combos:if x[int(i[0])] == x[int(i[1])] == x[int(i[2])] and x[int(i[0])] in "XO":return x[int(i[0])]return "D"

大神解答 NO.3

# migrated from python 2.7 def checkio(game_result):sample = "".join(game_result)data = game_result + [sample[i:9:3] for i in range(3)] + [sample[0:9:4], sample[2:8:2]]if "OOO" in data:return "O"elif "XXX" in data:return "X"else:return "D"

大神解答 NO.4

def checkio(game_result):patterns = [] + game_resultsize = len(game_result)for col in range(size):patterns.append(''.join([game_result[row][col] for row in range(size)]))patterns.append(''.join([game_result[x][x] for x in range(size)]))patterns.append(''.join([game_result[x][size - x - 1] for x in range(size)]))return 'X' if 'XXX' in patterns else 'O' if 'OOO' in patterns else 'D'

大神解答 NO.5

def checkio(game_result):result = 'D'if game_result[0][0] == game_result[1][1] == game_result[2][2] and game_result[0][0] != '.':result = game_result[0][0]return resultif game_result[2][0] == game_result[1][1] == game_result[0][2] and game_result[2][0] != '.':result = game_result[2][0]return result for i in range(3):if game_result[i][0] == game_result[i][1] == game_result[i][2] and game_result[i][0] != '.':result = game_result[i][0]breakif game_result[0][i] == game_result[1][i] == game_result[2][i] and game_result[0][i] != '.':result = game_result[0][i]breakreturn result 創作挑戰賽新人創作獎勵來咯，堅持創作打卡瓜分現金大獎

總結

以上是生活随笔為你收集整理的【Python CheckiO 题解】Xs and Os Referee的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。