LeetCode August check-in day11 130. Surrounded area (DFS)

Topic link

Ideas:

• We can perform DFS on the'O' on the boundary first, and modify all the'O's to other characters first. Here the'Y' character is taken.
• After searching for the four edges, the remaining'O' is changed to an'X' character, and we then change the'Y' character to an'O' character to get the correct answer.

Code:

class Solution {
    
    
public:
    int dx[4]={
    
    0,0,1,-1};
    int dy[4]={
    
    1,-1,0,0}; 
    void dfs(int x,int y,vector<vector<char>>& board) 
    {
    
    
        board[x][y]='Y';
        for(int i=0;i<4;i++)
        {
    
    
            int xx=x+dx[i];
            int yy=y+dy[i];
            if(xx>=1&&xx<board.size()-1&&yy>=1&&yy<board[0].size()-1&&board[xx][yy]=='O')
            {
    
    
                dfs(xx,yy,board);
            }
        }
        return ;
    }
    void solve(vector<vector<char>>& board) {
    
    
        if(board.size()!=0)
       {
    
    
           int n=board.size();
           int m=board[0].size();
           for(int i=0;i<n;i++)
           {
    
    
               if(board[i][0]=='O')
               dfs(i,0,board);
               if(board[i][m-1]=='O')
               dfs(i,m-1,board);
           }
             for(int i=0;i<m;i++)
           {
    
    
                if(board[0][i]=='O')
               dfs(0,i,board);
               if(board[n-1][i]=='O')
               dfs(n-1,i,board);
           }
         for(int i=0;i<board.size();i++)
        {
    
    
            for(int j=0;j<board[i].size();j++)
            {
    
    
                if(board[i][j]=='Y')
                board[i][j]='O';
                else if(board[i][j]=='O')
                board[i][j]='X';
            }
        }
       }
    }
};