LeetCode: Surrounded Regions [130]

class Solution {
public:

    void O2M(vector<vector<char> >&board, int i, int j){
        //从board[i][j]开始，遍历'O'区域，把'O'替换成'M'
        int rows=board.size();
        int cols=board[0].size();
        queue<int> qe;
        qe.push(i*cols+j);
        board[i][j]='M';
        while(!qe.empty()){
            int pos = qe.front(); qe.pop();
            i=pos/cols;
            j=pos%cols;
            //判断上方
            if(i-1>=0 && board[i-1][j]=='O'){
                board[i-1][j]='M';  //注意在将相邻元素填到队列中时，需要将它标记为以访问，否则以后在确定其他节点的'O'相邻位置时，很有可能又把这个节点加入到队列中。造成死循环。
                qe.push((i-1)*cols + j);
            }
            //判断下方
            if(i+1<rows && board[i+1][j]=='O'){
                board[i+1][j]='M';
                qe.push((i+1)*cols + j);
            }
            //判断左方
            if(j-1>=0 && board[i][j-1]=='O'){
                board[i][j-1]='M';
                qe.push(i*cols + j-1);
            }
            //判断右方
            if(j+1<cols && board[i][j+1]=='O'){
                board[i][j+1]='M';
                qe.push(i*cols + j+1);
            }
        }
    }

    void solve(vector<vector<char>> &board) {
        int rows=board.size();
        if(rows==0)return;
        int cols=board[0].size();
        if(cols==0)return;
        
        //把临边的'O'区域替换成'M'
        //上边
        for(int j=0; j<cols; j++){
            if(board[0][j]=='O')O2M(board, 0, j);
        }
        //下边
        for(int j=0; j<cols; j++){
            if(board[rows-1][j]=='O')O2M(board, rows-1, j);
        }
        //左边
        for(int i=0; i<rows; i++){
            if(board[i][0]=='O')O2M(board, i, 0);
        }
        //右边
        for(int i=0; i<rows; i++){
            if(board[i][cols-1]=='O')O2M(board, i, cols-1);
        }
        
        //扫描矩阵，把O替换成X, 把M替换成O
        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]=='M')board[i][j]='O';
            }
        }
        
    }
};