問題
Minimax is how game-playing programs choose moves: assume both players play perfectly, score every finished game, and back the scores up the game tree. The player to move picks the child with the highest score for itself.
The board is a 9-character string read row by row (indices 0-8), with "X", "O" or "." for an empty cell. X always moves first, so it is X's turn when both have the same number of marks and O's turn otherwise.
Score a finished game from the point of view of the player to move: +1 for a win, 0 for a draw, -1 for a loss. Return the empty cell whose move has the best minimax score; if several are equally good, return the lowest index.
Implement best_move(board).
Input
board: A valid position where nobody has won yet and at least one cell is empty.
Output
Return the index (0-8) of the best move.
制約
board has exactly 9 characters, each X, O or .
The position is reachable in a real game, nobody has won, and at least one cell is empty
例
入力：
board = "XX.OO...."
期待される出力：
expected = 2入力：
std::string board = "XX.OO....";
期待される出力：
long long expected = 2;入力：
string board = "XX.OO....";
期待される出力：
long expected = 2;解説
X and O have two marks each, so X moves. Playing 2 completes the top row and wins at once (+1).
入力：
board = "X..OO...X"
期待される出力：
expected = 5入力：
std::string board = "X..OO...X";
期待される出力：
long long expected = 5;入力：
string board = "X..OO...X";
期待される出力：
long expected = 5;解説
X moves. O threatens to complete the middle row at 5. Every other move lets O win (-1), while blocking at 5 leads to a draw (0), so the answer is 5.
解き方のヒント
まずは自分で考えてから、ヒントを 1 つずつ開いてみましょう。
ヒント 1
Determine whose turn it is from the counts of X and O, then evaluate every legal move recursively.
ヒント 2
Score terminal boards from the current player’s perspective, and choose the move that maximizes that value.
ヒント 3
If multiple moves tie, keep the earliest empty cell index, and ensure wins are detected before considering further moves.
目安の計算量
O(9!) time, O(9) extra space
参考資料
この課題はCodeUtility用に調整されています。上記の説明と制約に従ってください。