python 给出一个由‘O’和‘X’组成的矩阵，在矩阵中的所有‘X’所在的行、列都需要变成‘O’。 如果在矩阵边界上的‘O’所在的行、列被‘X’包围，则这些‘O’也需要变成‘X’。 样例： 输入： [ ["X","X","X","X"], ["X","O","O","X"], ["X","X","O","X"], ["X","O","X","X"] ] 输出： [ ["X","X","X","X"], ["X","X","X","X"], ["X","X","X","X"], ["X","O","X","X"] ]

可以使用DFS或BFS来解决此问题。下面是使用DFS的Python代码：

class Solution:
    def solve(self, board: List[List[str]]) -> None:
        """
        Do not return anything, modify board in-place instead.
        """
        if not board:
            return
        
        m, n = len(board), len(board[0])
        
        # DFS遍历边界
        def dfs(i, j):
            if i < 0 or i >= m or j < 0 or j >= n or board[i][j] != 'O':
                return
            board[i][j] = '#'
            dfs(i-1, j)  # 上
            dfs(i+1, j)  # 下
            dfs(i, j-1)  # 左
            dfs(i, j+1)  # 右
        
        for i in range(m):
            dfs(i, 0)
            dfs(i, n-1)
        for j in range(n):
            dfs(0, j)
            dfs(m-1, j)
        
        # 标记'O'和'X'
        for i in range(m):
            for j in range(n):
                if board[i][j] == 'O':
                    board[i][j] = 'X'
                elif board[i][j] == '#':
                    board[i][j] = 'O'

注意，这里使用了Python的列表和递归函数，需要对输入进行检查以避免出现越界等错误。