class Solution {
public:
    int rows, columns;
    void solve(vector<vector<char>>& board) {
        rows = board.size();
        columns = board[0].size();
        for(auto i = 0; i < rows; ++i){
            dfs(board, i, 0);   //左边界
            dfs(board, i, columns - 1); //右边界
        }
        for(auto j = 0; j < columns; ++j){
            dfs(board, 0, j);   //上边界
            dfs(board, rows - 1, j);    //下边界
        }
        for(auto i = 0; i < rows; ++i){
            for(auto j = 0; j < columns; ++j){
                if(board[i][j] == 'A')
                    board[i][j] = 'O';
                else if(board[i][j] == 'O')
                    board[i][j] = 'X';
            }
        }
    }
    void dfs(vector<vector<char>>& board, int row, int column){
        //如果超过边界或者不是‘O’
        //只有包含边界上的O的区域才会被保留，先把存在边界上的O的所有区域改为A
        //那么剩下的O就是包裹在内部的，也就是要改为X的
        if(row < 0 || row >= rows || column < 0 || column >= columns || board[row][column] != 'O'){
            return;
        }
        board[row][column] = 'A';   //和边界上的O接触过的都先换成A
        dfs(board, row - 1, column);    //上
        dfs(board, row, column - 1);    //左
        dfs(board, row + 1, column);    //下
        dfs(board, row, column + 1);    //右
    }
};

Accepted
58/58 cases passed (12 ms)
Your runtime beats 69.35 % of cpp submissions
Your memory usage beats 58.44 % of cpp submissions (9.8 MB)