【surrounded-regions】

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

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

For example,

X X X XX O O XX X O XX O X X

After running your function, the board should be:

X X X XX X X XX X X XX O X X

题意：O周围都换成x，这个题目的关注点是在四条边的0是不能替换的。

思路：

DFS深度优先：

从外围是‘0’的开始深度往里走，这时候里面的'0'就有两种命运，一种是跟外围的'0'是联通的，那么这些‘0’就可以存活，剩下的孤立

的‘0’就没办法了，就只能被‘X’

class Solution{public:void dfs(int row, int col, vector<vector<char>>&  board){if (row<0 || row>=rows || col<0 || col>=cols){return;}if (board[row][col]!='O'){return;}board[row][col] = '+';dfs(row - 1, col, board); //updfs(row, col + 1, board); //rightdfs(row + 1, col, board); //downdfs(row, col - 1, board); //left}void solve(vector<vector<char>>& board){if (board.size()==0 || board[0].size()==0){return;}rows = board.size();cols = board.size();for (int i=0; i<rows; i++){dfs(i, 0, board);//第一列dfs(i, cols-1, board);//最后一列}for (int j=0; j<cols; ++j){dfs(0, j, board);//第一行dfs(rows-1, j, board);//最后一行}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]=='+'){board[i][j] = 'O';}}}}private:int rows;//总行号int cols;//总列号};