130. Surrounded Regions

最新推荐文章于 2025-12-06 07:43:37 发布

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#algorithms #Dynamic Programming #leetcode

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本文介绍了一种解决二维棋盘中被'X'包围的'O'区域问题的算法实现。通过深度优先搜索从棋盘边界开始标记可能不受包围的'O'，最终将未被标记的'O'转变为'X'，实现区域围捕。

problem:

Given a 2D board containing 'X' and 'O' (the letter 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

solution:

class Solution {
public:
	void solve(vector<vector<char>>& board) {
        int i,j;
        int row=board.size();
        if(!row)
        	return;
        int col=board[0].size();

		for(i=0;i<row;i++){
			check(board,i,0,row,col);
			if(col>1)
				check(board,i,col-1,row,col);
		}
		for(j=1;j+1<col;j++){
			check(board,0,j,row,col);
			if(row>1)
				check(board,row-1,j,row,col);
		}
		for(i=0;i<row;i++)
			for(j=0;j<col;j++)
				if(board[i][j]=='O')
					board[i][j]='X';
		for(i=0;i<row;i++)
			for(j=0;j<col;j++)
				if(board[i][j]=='1')
					board[i][j]='O';
    }
	void check(vector<vector<char> >&vec,int i,int j,int row,int col){
		if(vec[i][j]=='O'){
			vec[i][j]='1';
			if(i>1)
				check(vec,i-1,j,row,col);
			if(j>1)
				check(vec,i,j-1,row,col);
			if(i+1<row)
				check(vec,i+1,j,row,col);
			if(j+1<col)
				check(vec,i,j+1,row,col);
		}
	}
};