LeetCode: Surrounded Regions

思路：刚开始觉得如果每个'O'点如果四个方向上存在'X'点，则其被包围，其实如果存在一条路径，即所有'O'点相连，最后连接到边界上去了，那么这个点就没有被包围，所以，只能通过边界的'O'点来进行深度搜索，只要与边界'O'相连通的'O'点都是没有被覆盖的，剩下的就好办了。

code:

class Solution {
public:
    bool isRightLoc(int i,int j,int rows,int cols){
        if(i <rows && i >=0 && j < cols && j >=0)
            return true;
        return false;
    }
    void solve(vector<vector<char>> &board) {
        if(board.size() == 0)return;
        int rows = board.size(), cols = board[0].size();
        for(int j = 0;j < cols;j++){
            if(board[0][j] == 'O'){
               //board[0][j] = 'T';
				//cout<<"sdf"<<endl;
                dfs(0,j,board);
            }
			
            if(board[rows-1][j] == 'O'){
                //board[rows-1][j] = 'T';
                dfs(rows-1,j,board);
            }
        }
        for(int i = 1;i < rows-1;i++){
            if(board[i][0] == 'O'){
                //board[i][0] = 'T';
                dfs(i,0,board);
            }
            if(board[i][cols-1] == 'O'){
                //board[i][cols-1] = 'T';
                dfs(i,cols-1,board);
            }
        }
        
        for(int i = 0;i < rows;i++)
            for(int j = 0;j < cols;j++){
                if(board[i][j] == 'O')
                    board[i][j] = 'X';
                else if(board[i][j] == 'T')
                    board[i][j] = 'O';
            }
    }
    void dfs(int i,int j,vector<vector<char>> &board){
		
        int rows = board.size(), cols = board[0].size();
        if(isRightLoc(i,j,rows,cols)){
			//cout<<i<<" "<<j<<" "<<board[i][j]<<endl;
            if(board[i][j] == 'O'){
                board[i][j] = 'T';
                dfs(i,j+1,board);
                dfs(i-1,j,board);
                dfs(i,j-1,board);
                dfs(i+1,j,board);
            }
        }
    }
};