题解 | #牛群的活动区域#

class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param board char字符型vector<vector<>> 
     * @return char字符型vector<vector<>>
     */
    vector<vector<int>> directions {{-1, 0}, {0, 1}, {1, 0}, {0, -1}};
    // 先将和边缘的B相连的所有B全部变为#
    void dfs(vector<vector<char>>& board,int row, int col,  int x, int y) {
        board[x][y] = '#';
        for (int i = 0; i < 4; i++) {
            int new_x = directions[i][0];
            int new_y = directions[i][1];
            if (new_x >= 0 && new_x < row && new_y >= 0 && new_y < col && board[new_x][new_y] == 'B') {
                dfs(board, row, col, new_x, new_y);
            }
        }
    }
    vector<vector<char> > solve(vector<vector<char> >& board) {
        // write code here
        int row = board.size(), col = board[0].size();
	  //先将边缘的B和与之相连的B全部变为#
        for (int i = 0; i < row; i++) {
            if (board[i][0] == 'B') {
                dfs(board, row, col, i, 0);
            }
            if (board[i][col-1] == 'B') {
                dfs(board, row, col, i, col-1);
            }
        }
	  
        for (int j = 1; j < col-1; j++) {
            if (board[0][j] == 'B') {
                dfs(board, row, col, 0, j);
            }
            if (board[row-1][j] == 'B') {
                dfs(board, row, col, row-1, j);
            }
        }
	  //再遍历整个矩阵，如果为B，则改为A，如果为# 则改为B,注意顺序不能颠倒。
        for (int i = 0; i < row; i++) {
            for (int j = 0; j < col; j++) {
                
                if (board[i][j] == 'B') {
                    board[i][j] = 'A';
                }
                if (board[i][j] == '#') {
                    board[i][j] = 'B';
                }
            }
        }
        return board;
    }
};