[leetcode] Surrounded Regions

最新推荐文章于 2020-08-11 10:37:01 发布

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本博客详细介绍了如何通过深度优先搜索（DFS）算法来解决2D棋盘中被‘X’包围的‘O’区域问题。通过对边界‘O’进行标记，然后翻转所有未被标记的‘O’为‘X’，实现区域的捕获。代码示例清晰展示了实现过程。

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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

class Solution {
int m,n;
public:
    void solve(vector<vector<char> > &board) {
    	// Start typing your C/C++ solution below
		// DO NOT write int main() function
		m=board.size();
        if(m==0)
            return;
		n=board[0].size();
		for(int i=0 ; i<n ; i++){
			DFS(board,0,i);
			DFS(board,m-1,i);
		}
		for(int i=1 ; i<m-1 ; i++){
			DFS(board,i,0);
			DFS(board,i,n-1);
		}
		for(int i=0 ; i<m ; i++){
			for(int j=0 ; j<n ; j++){
				if(board[i][j]=='O')
					board[i][j]='X';
				if(board[i][j]=='\0')
					board[i][j]='O';
			}
		}
	}
	void DFS(vector<vector<char> >&board, int i, int j){
		if(i<0 || j<0 || i>=m || j>=n || board[i][j]=='X'||board[i][j]=='\0')
			return;
		board[i][j]='\0';
		DFS(board,i+1,j);
		DFS(board,i-1,j);
		DFS(board,i,j+1);
		DFS(board,i,j-1);
	}
};