Python递归-矩阵中子序列的最大长度未更新

我试图找到矩阵中字符的最长序列。我对python真的很陌生，我认为问题在于python的递归方法与C/C++/Java等不同。这是我的代码，现在。。。如果在Python中不使用递归，您知道另一种方法吗？或者您可以修复我的代码以在pythonic递归中工作吗？（问题是长度和访问矩阵在递归过程中没有更新）

我想这个就行了

def get_current_score(self, i, j, char, leng):
    # leng = 0 <- don't
    self.viz[i][j] = True  # you set it twice, let's keep that one

    def test_cell(a, b):  # keep it DRY
        return self.out_of_span(a, b) == False \
               and self.viz[a][b] == False \
               and self.matrix[a][b] == char

    cells = [(i - 1, j), (i - 1, j + 1), (i - 1, j - 1), (i, j - 1),
             (i, j + 1), (i + 1, j), (i + 1, j - 1), (i + 1, j + 1)]
    for a, b in cells:
        if test_cell(a, b): 
            # you don't need to set that cell to true since that's the 
            # first thing you do in the function
            # self.viz[a][b] = True
            return leng + self.get_current_score(a, b, char, leng)

    return 1 + leng

def add(self, index):
    [...]
    # scor
    print('\n --------------\n')
    for i in range(self.maxh, self.nr):
        for j in range(self.span[0], self.span[1]+1):
                if(self.player1 == False and self.matrix[i][j] == 'X'):
                    # no need to send self.viz here. same for score2
                    # if you need to set self.score1 to 0 do it here. not in the recursion
                    self.score1 = max(self.score1, self.get_current_score(i, j, 'X', self.score1))
                    self.viz[i][j] = True
                    #self.score1 -= 1
                else:
                    if(self.player1 == True and self.matrix[i][j] == 'O'):
                        self.score2 = max(self.score2, self.get_current_score(i, j, 'O', self.score1))
                        self.viz[i][j] = True
    self.reset_viz()
    self.print_matrix()

def获取当前分数（self、i、j、char、leng）：
#leng=0编辑：这不是最终代码，抱歉，实际代码使用leng=而不是返回leng+，不清楚此代码试图实现什么，您能提供更多信息吗？另外，如何定义子序列？（查看代码，似乎您没有使用通用的“子序列”定义）这很难闻：您将类属性
self.viz
作为参数
viz
传递，并且只使用
self.viz
。对于更糟糕的长度：发送
self.score1
将其设置为局部变量
leng
，该变量在每次递归时设置为零。您的函数只能返回1
def get_current_score(self, i, j, char, leng):
    # leng = 0 <- don't
    self.viz[i][j] = True  # you set it twice, let's keep that one

    def test_cell(a, b):  # keep it DRY
        return self.out_of_span(a, b) == False \
               and self.viz[a][b] == False \
               and self.matrix[a][b] == char

    cells = [(i - 1, j), (i - 1, j + 1), (i - 1, j - 1), (i, j - 1),
             (i, j + 1), (i + 1, j), (i + 1, j - 1), (i + 1, j + 1)]
    for a, b in cells:
        if test_cell(a, b): 
            # you don't need to set that cell to true since that's the 
            # first thing you do in the function
            # self.viz[a][b] = True
            return leng + self.get_current_score(a, b, char, leng)

    return 1 + leng

def add(self, index):
    [...]
    # scor
    print('\n --------------\n')
    for i in range(self.maxh, self.nr):
        for j in range(self.span[0], self.span[1]+1):
                if(self.player1 == False and self.matrix[i][j] == 'X'):
                    # no need to send self.viz here. same for score2
                    # if you need to set self.score1 to 0 do it here. not in the recursion
                    self.score1 = max(self.score1, self.get_current_score(i, j, 'X', self.score1))
                    self.viz[i][j] = True
                    #self.score1 -= 1
                else:
                    if(self.player1 == True and self.matrix[i][j] == 'O'):
                        self.score2 = max(self.score2, self.get_current_score(i, j, 'O', self.score1))
                        self.viz[i][j] = True
    self.reset_viz()
    self.print_matrix()