【Leetcode】130. Surrounded Regions（DFS）

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Given a 2D board containing 'X' and 'O' (the letter O), capture all regions surrounded by 'X'.

A region is captured by flipping all 'O's into 'X's in that surrounded region.

Example:

X X X X
X O O X
X X O X
X O X X

After running your function, the board should be:

X X X X
X X X X
X X X X
X O X X

题目大意：

二维表中，将被X包围的O全部转换成X。

解题思路：

那么其中满足条件的只可能是边界上或者与边界相连的O能够幸免于难。

所以每次找到边界上等于O时，DFS将其和与其相连的都标记为#。

class Solution {
private:
    int x[4]={-1,1,0,0};
    int y[4]={0,0,1,-1};
    vector<vector<int>> begin_point;
    bool valid(vector<vector<char> >& board, int cor_x, int cor_y){
        if(cor_x>=0&&cor_x<board.size()&&cor_y>=0&&cor_y<board[0].size()&&board[cor_x][cor_y]=='O'){
            return true;
        }
        return false;
    }
    void dfs(vector<vector<char> >& board, int cor_x, int cor_y){
        
        if(board[cor_x][cor_y]=='O'){
            board[cor_x][cor_y] = '#';
        }
        
        for(int i=0;i<4;i++){
            int xx = cor_x + x[i];
            int yy = cor_y + y[i];
            if(valid(board, xx, yy)){
                dfs(board, xx, yy);
            }
        }
        
    }
    
public:
    void solve(vector<vector<char>>& board) {
        if(board.size()==0) return;
        for(int i=0;i<board.size();i++){
            for(int j=0;j<board[0].size();j++){
                
                if(i==0||i==board.size()-1||j==0||j==board[0].size()-1){
                    if(board[i][j]=='O'){
                        // vector<int> tmp;
                        // tmp.push_back(i);
                        // tmp.push_back(j);
                        dfs(board, i, j);
                    }
                    // begin_point.push_back(tmp);
                }
            }
        }  
        
        for(int i=0;i<board.size();i++){
            for(int j=0;j<board[0].size();j++){
                if(board[i][j]=='O'){
                    board[i][j]='X';
                }
                else if(board[i][j]=='#'){
                    board[i][j]='O';
                }
            }
        } 
        
    }
};