from typing import List
# 這道題看了大佬寫的程式碼，經過自己的理解寫出來了。
# 從最外圍的四周找有沒有為O的，如果有的話就進入深搜函式，然後深搜遍歷
# 判斷上下左右的位置是否為O
class Solution:
    def solve(self, board: List[List[str]]) -> None:
        # 判斷是否為空列表
        if not board or not board[0] : return
        # 求出列表的有幾行和有幾列
        row = len(board)
        col = len(board[0])
 
        # 深搜，當邊界為O進行判斷
        def dfs(index1,index2):
            # 首先將O變成B，之後再變回來。
            board[index1][index2] = 'B'
            # 遍歷上下左右
            for x,y in [(0,1),(0,-1),(1,0),(-1,0)]:
                print(x,y)
                # 注意這裡不能改變index1和index2的值，因為後邊還要用。
                i = index1 +  x
                j = index2 +  y
 
                # 如果當前位置的上下左右不出邊界而且位置上為O就，進行深搜
                # 注意這裡邊界上邊不能判斷，
                if 1 <= i < row and 1 <= j < col and board[i][j] == "O":
                    print("111")
                    dfs(i,j)
        # 判斷第一行和最後一行上的邊界值
        for index1 in range(col):
            if board[0][index1] == "O":
 
                dfs(0,index1)
            if board[row - 1][index1] == 'O':
                dfs(row - 1,index1)
        # 判斷第一列和最後一列的邊界值
        for index2 in range(row):
            if board[index2][0] == "O":
                dfs(index2,0)
            if board[index2][col - 1] == 'O':
                dfs(index2,col - 1)
        # 將O全部改寫成X，將之前寫的B也改寫成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"]])
)