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 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:从四条边上的'0'入手,与其邻接的'O'都不被改,记为'D'。逆向思维!从不被改的'O'入手,而非寻找被改得'O'

class Solution {
public:
void dfs(int x, int y){
if(x< || x>=m || y< || y>=n || board[x][y]!='O') return;
board[x][y]='D'; //图的四个方向遍历
dfs(x-,y);
dfs(x+,y);
dfs(x,y-);
dfs(x,y+);
} void solve(vector<vector<char>> &board){
if (board.empty()) return;
this->board = board;
m=board.size();
n=board[].size();
if(n<= || m<=) return; for(int j=;j<n;j++){
dfs(,j);
dfs(m-,j);
} for(int i=;i<m;i++){
dfs(i,);
dfs(i,n-);
} for(int i=;i<m;i++)
for(int j=;j<n;j++){
if(this->board[i][j]=='O') this->board[i][j]='X';
else if(this->board[i][j]=='D') this->board[i][j]='O';
}
board = this->board;
}
private:
int m,n;
vector<vector<char>> board;
};

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