被包围的棋子 Surrounded Regions

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

思路：目标是要找到由X包围起来的O的区域。

首先，外围一圈上的O肯定会保留下来。然后，从外围的O能达到的O也要保留。剩下其他的O就是内部的O。

所以方法就是从外围的一圈进行DFS算法：依次对外圈的“O”做DFS，将其可以到达O临时设置为#。

特殊用例：只有外围轮廓没有内部。比如长或者宽小于2，此时不存在被包围的'X'。

class Solution {
public:
    void solve(vector<vector<char>> &board) {
        int row = board.size();
        if(row < 2)
            return;
        int col = board[0].size();
        if(col < 2)
            return;
        
        //从最外面的一圈进行dfs
        //top一行
        for(int i=0;i<col;i++)
        {
            if(board[0][i] == 'O')
            {
                board[0][i] = '#';
                dfs(board, 0, i, row, col);
            }
        }
        // bottom一行
        for(int i=0;i<col;i++)
        {
            if(board[row-1][i] == 'O')
            {
                board[row-1][i] = '#';
                dfs(board, row-1, i, row, col);
            }
        }
        //left一列
        for(int i=1;i<row-1;i++)
        {
            if(board[i][0] == 'O')
            {
                board[i][0] = '#';
                dfs(board, i, 0, row, col);
            }
        }
        // right一列
        for(int i=1;i<row-1;i++)
        {
            if(board[i][col-1] == 'O')
            {
                board[i][col-1] = '#';
                dfs(board, i, col-1, row, col);
            }
        }
        
        //将'O'变为'X'，将'#'恢复回'O'
        for(int i=0;i<row;i++)
            for(int j=0;j<col;j++)
            {
                if(board[i][j] == 'O')
                    board[i][j] = 'X';
                else if(board[i][j] == '#')
                    board[i][j] = 'O';
            }
    }
    
    void dfs(vector<vector<char>> &board, int i, int j, int row, int col)
    {
        if(i > 1 && board[i-1][j] == 'O')
        {
            board[i-1][j] = '#';
            dfs(board, i-1, j, row, col);
        }
        if(i < row-1 && board[i+1][j] == 'O')
        {
            board[i+1][j] = '#';
            dfs(board, i+1, j, row, col);
        }
        if(j > 1 && board[i][j-1] == 'O')
        {
            board[i][j-1] = '#';
            dfs(board, i, j-1, row, col);
        }
        if(j < col-1 && board[i][j+1] == 'O')
        {
            board[i][j+1] = '#';
            dfs(board, i, j+1, row, col);
        }
    }
};