130被围绕的区域

from typing import List
# 这道题看了大佬写的代码，经过自己的理解写出来了。
# 从最外围的四周找有没有为O的，如果有的话就进入深搜函数，然后深搜遍历
# 判断上下左右的位置是否为O
class Solution:
    def solve(self, board: List[List[str]]) -> None:
        # 判断是否为空列表
        if not board or not board[0] : return
        # 求出列表的有几行和有几列
        row = len(board)
        col = len(board[0])
        # 深搜，当边界为O进行判断
        def dfs(index1,index2):
            # 首先将O变成B，之后再变回来。
            board[index1][index2] = 'B'
            # 遍历上下左右
            for x,y in [(0,1),(0,-1),(1,0),(-1,0)]:
                print(x,y)
                # 注意这里不能改变index1和index2的值，因为后边还要用。
                i = index1 +  x
                j = index2 +  y
                # 如果当前位置的上下左右不出边界而且位置上为O就，进行深搜
                # 注意这里边界上边不能判断，
                if 1 <= i < row and 1 <= j < col and board[i][j] == "O":
                    print("111")
                    dfs(i,j)
        # 判断第一行和最后一行上的边界值
        for index1 in range(col):
            if board[0][index1] == "O":
                dfs(0,index1)
            if board[row - 1][index1] == 'O':
                dfs(row - 1,index1)
        # 判断第一列和最后一列的边界值
        for index2 in range(row):
            if board[index2][0] == "O":
                dfs(index2,0)
            if board[index2][col - 1] == 'O':
                dfs(index2,col - 1)
        # 将O全部改写成X，将之前写的B也改写成O
        for index1 in range(row):
            for index2 in range(col):
                if board[index1][index2] == "O":
                    board[index1][index2] = "X"
                if board[index1][index2] == "B":
                    board[index1][index2] = 'O'

A = Solution()
print(A.solve([["O","X","X","O","X"],["X","O","O","X","O"],["X","O","X","O","X"],["O","X","O","O","O"],["X","X","O","X","O"]])
)