【每周一坑】罗马数字转换

def romanToInt(s):
    # your code here
    return i

assert romanToInt('III') == 3
assert romanToInt('IV') == 4
assert romanToInt('VI') == 6
assert romanToInt('XIX') == 19
assert romanToInt('XLV') == 45
assert romanToInt('MCMLXXX') == 1980
assert romanToInt('CMXCIX') == 999

class Spiral:
    def __init__(self, N):
        # 构造一个二维数组
        self.matrix = [[None for i in range(N)] for j in range(N)]
        # 起始行列数
        self.row = 0
        self.col = 0
        # 数组的边界值
        self.max_row = N
        # 更换方向的标记
        self.mark = 0

    # 按需取出数组运动方向
    def derection(self, mark):
        around = [
            [self.row, self.col+1], # 向右
            [self.row+1, self.col], # 向下
            [self.row, self.col-1], # 向左
            [self.row-1, self.col]  # 向上
        ]
        return around[mark%4]

    # 针对目前位置，获取下一位置的行列数
    # 下一位置为边界则更换方向
    # 下一位置已经有元素则更换方向
    def next(self):
        # 下一位置
        i = self.derection(mark=self.mark)
        # 判断是否更换方向，不更换则更新 self.row / self.col
        if -1 not in i and self.max_row not in i:
            if self.matrix[i[0]][i[1]] is None:
                self.row,self.col = i
                return None
        # 更换方向
        self.mark += 1
        return self.next()

    def solution(self):
        # 逐一取出 1 到 n^2 值
        for i in range(1,self.max_row**2+1):
            # 按行列赋值
            self.matrix[self.row][self.col] = i
            # 退出条件
            if i == self.max_row**2:
                break
            # 更新行列值
            self.next()
        # 打印结果
        for r in self.matrix:
            for c in r:
                print('{0:^{1}}'.format(c,self.max_row), end=' ')
            print('\n')

if __name__ == '__main__':
    n = int(input('>>>'))
    s = Spiral(n)
    s.solution()