[leetcode]Design Tic-Tac-Toe

Posted 2021-03-11 阿牧遥

正确做法是: 只保存行/列/对角线的和，而不用保存所有元素，空间复杂度从O(n2)降到O(n)；move()只需判断四个值是否 = n - 1: 当前行sum，当前列sum，两个对角线，时间复杂度为O(1)

而我只是用dict记录是否还有可能赢。差强人意。

class TicTacToe:

    def __init__(self, n: int):
        """
        Initialize your data structure here.
        """
        self.board = [[0] * n for i in range(n)]
        self.validRows = set([i for i in range(n)])
        self.validCols = set([i for i in range(n)])
        self.validCrosses = set([0, 1])

    def move(self, row: int, col: int, player: int) -> int:
        """
        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.
        """
        n = len(self.board)
        
        self.board[row][col] = player
        # each row
        for i in range(n):
            if i not in self.validRows:
                continue
            cnt1 = 0
            cnt2 = 0
            for j in range(n):
                if self.board[i][j] == 1:
                    cnt1 += 1
                elif self.board[i][j] == 2:
                    cnt2 += 1
            if cnt1 == n:
                return 1
            elif cnt2 == n:
                return 2
            if cnt1 > 0 and cnt2 > 0:
                self.validRows.remove(i)
        # each column
        for i in range(n):
            if i not in self.validCols:
                continue
            cnt1 = 0
            cnt2 = 0
            for j in range(n):
                if self.board[j][i] == 1:
                    cnt1 += 1
                elif self.board[j][i] == 2:
                    cnt2 += 1
            if cnt1 == n:
                return 1
            elif cnt2 == n:
                return 2
            if cnt1 > 0 and cnt2 > 0:
                self.validCols.remove(i)
        # two cross
        if 0 in self.validCrosses:
            cnt1 = 0
            cnt2 = 0
            for i in range(n):
                if self.board[i][i] == 1:
                    cnt1 += 1
                elif self.board[i][i] == 2:
                    cnt2 += 1
            if cnt1 == n:
                return 1
            elif cnt2 == n:
                return 2
            if cnt1 > 0 and cnt2 > 0:
                self.validCrosses.remove(0)
        if 1 in self.validCrosses:
            cnt1 = 0
            cnt2 = 0
            for i in range(n):
                if self.board[i][n-i-1] == 1:
                    cnt1 += 1
                elif self.board[i][n-i-1] == 2:
                    cnt2 += 1
            if cnt1 == n:
                return 1
            elif cnt2 == n:
                return 2
            if cnt1 > 0 and cnt2 > 0:
                self.validCrosses.remove(1)
        
        return 0
                    
                
        


# Your TicTacToe object will be instantiated and called as such:
# obj = TicTacToe(n)
# param_1 = obj.move(row,col,player)