题目：被围绕的区域

 

思路：

好吧，这题不会。

bfs+递归 dfs+非递归 dfs+并查集 - 被围绕的区域 - 力扣（LeetCode）

 将问题转化为与边界O相连的O。有点像岛屿问题了。

 

代码是：

//code
 
class Solution {
public:
    void dfs(vector<vector<char>>& board , int i, int j){
        if(i<0 || j<0 || i >=board.size() || j >= board[0].size() || board[i][j]=='X' || board[i][j]=='#'){
            return ; 
        }
        board[i][j]='#';
        dfs(board,i-1,j);
        dfs(board,i+1,j);
        dfs(board,i,j-1);
        dfs(board,i,j+1);
    }
    void solve(vector<vector<char>>& board) {
        //if(board.size()==0) return;
        int row = board.size();
        int col = board[0].size();
        for(int i=0;i<row;i++){
            for(int j=0;j<col;j++){
                bool isedge = (i==0 || j == 0 || i==row-1 || j==col-1);
                if(isedge && board[i][j]=='O'){
                    dfs(board,i,j);
                }
            }
        }
        for(int i=0;i<row;i++){
            for(int j = 0;j<col;j++){
                switch(board[i][j]){
                    case('#'):
                        board[i][j]='O';
                        break;
                    case('O'):
                        board[i][j]='X';
                        break;
                }
            }
        }
    }
};