leetcode Surrounded Regions

Do_Cool_Thing

于 2014-08-08 16:02:46 发布

阅读量404

点赞数

CC 4.0 BY-SA版权

分类专栏： leetcode Fighting

本文链接：https://blog.youkuaiyun.com/menyangyang/article/details/38440649

leetcode Fighting 专栏收录该内容

133 篇文章

订阅专栏

本文介绍了一种使用深度优先搜索(DFS)算法解决特定问题的方法：将二维矩阵中被'X'包围的'O'替换为'X'，而处于边界的'O'所邻接的'O'则不被替换。通过从边界开始遍历，可以有效识别并处理矩阵中的'O'和'X'，确保边界'O'不受影响。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目要求的是将被'X'包围的'O'替换为‘X’，则处于边界的‘O’所邻接的'O'均不能被’X‘包围，可以从边界处进行DFS。

代码

class Solution
{
public:
	void solve(vector<vector<char>> &board)
	{
		if(board.empty()||board.size()==0||board[0].size()==0)
			return;

		int row = board.size();
		int col = board[0].size();	
		

		for(int i = 0; i < col; ++i)
		{
			judge4NearNeighbor( board,0,i);
			judge4NearNeighbor( board, row-1,i);		
		}


		for(int j = 0; j < row; ++j)
		{
			judge4NearNeighbor( board,j,0);
			judge4NearNeighbor( board, j,col-1);
		
		}

		for(int i = 0; i < row; ++i )
			for(int j = 0; j < col; ++j)
				if(board[i][j]=='Y')
					board[i][j] = 'O';
				else
					board[i][j] = 'X';

	}

	void judge4NearNeighbor(vector<vector<char>> &board, int i, int j)
	{
		if(i>=0&&i<=board.size()-1&&j>=0&&j<=board[0].size()-1&&board[i][j]=='O')
		{
			board[i][j] = 'Y';
			judge4NearNeighbor(board, i, j-1);
			judge4NearNeighbor(board, i-1, j);
			judge4NearNeighbor(board, i, j+1);
			judge4NearNeighbor(board, i+1, j);
		}
	

	}

};