[leetcode 130] Surrounded Regions

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此博客介绍了如何通过遍历二维数组并利用队列进行广度优先搜索来解决翻转被周围字符'X'包围的'O'字符的问题。详细解释了算法实现过程，并提供了代码示例。

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 {
public:
    void process(int i,int j,vector<vector<char> >& board)
	{
		int m=board.size();
		int n=board[0].size();

		typedef pair<int,int> point;
		queue<point> Q;
		Q.push(point(i,j));
		board[i][j]='E';
		while(!Q.empty())
		{
			point tmp=Q.front();
			Q.pop();
			
			int x=tmp.first,y=tmp.second;

			//extending
			if (x!=0&&board[x-1][y]=='O')
			{
				Q.push(point(x-1,y));
				board[x-1][y]='E';   //extended;
			}
			if (x!=m-1&&board[x+1][y]=='O')
			{
				Q.push(point(x+1,y));
				board[x+1][y]='E';   //extended;
			}
			if (y!=0&&board[x][y-1]=='O')
			{
				Q.push(point(x,y-1));
				board[x][y-1]='E';   //extended;
			}
			if (y!=n-1&&board[x][y+1]=='O')
			{
				Q.push(point(x,y+1));
				board[x][y+1]='E';   //extended;
			}
		}
	}
	void solve(vector<vector<char> >& board)
	{
		int m=board.size();
		if (m==0)
			return;
		int n=board[0].size();
		int i,j;
		for(i=0;i<m;i++)
		{
			if (board[i][0]=='O')
				process(i,0,board);
			if (board[i][n-1]=='O')
				process(i,n-1,board);
		}
		for(j=0;j<n;j++)
		{
			if (board[0][j]=='O')
				process(0,j,board);
			if (board[m-1][j]=='O')
				process(m-1,j,board);
		}
		for(i=0;i<m;i++)
			for(j=0;j<n;j++)
			{
			    board[i][j]=board[i][j]!='E'?'X':'O';
			}
	}	
};