Leetcode Surrounded Regions

最新推荐文章于 2025-08-08 00:00:33 发布

rosepicker

最新推荐文章于 2025-08-08 00:00:33 发布

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分类专栏： leetcode 文章标签： leetcode

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本文介绍了一种使用深度优先搜索(DFS)算法解决二维棋盘中被X包围的O型区域捕获问题的方法。通过首先将边缘上的O标记为Y，再将未被标记的O转换为X，最后将Y转换回O来实现。

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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

通过DFS将与边缘O直接相连的O设置为Y，然后在进行遍历board，将O设置为X，将Y设置为O

代码如下：

class Solution {
public:
    class point
    {
        public:
        int x;
        int y;
        point(int i,int j)
        {
            x=i;
            y=j;
        }
    };
    
    void solve(vector<vector<char>>& board) {
        if(board.empty())
            return;
        int m = board.size();
        int n = board[0].size();
        queue<point*> data;
       
        for(int i=0;i<m;i++)
            for(int j=0;j<n;j++)
            {
                if(j==0||j==n-1||i==0||i==m-1)
                {
                    if(board[i][j] == 'O')
                    {
                        board[i][j]  = 'Y';
                        data.push(new point(i,j));
                    }
                }
            }
        
        while(!data.empty())
        {
            point* now = data.front();
            data.pop();
            int x = now->x;
            int y = now->y;
            
            board[x][y]  = 'Y';
            
            if(x+1<m && board[x+1][y] == 'O' )
                data.push(new point(x+1,y));
            
            if(x-1>=0  && board[x-1][y] == 'O' )
                data.push(new point(x-1,y));
            
            if(y+1<n && board[x][y+1] == 'O' )
                data.push(new point(x,y+1));
            
            if(y-1>=0 && board[x][y-1] == 'O' )
                data.push(new point(x,y-1));
        }
        
        for(int i=0;i<m;i++)
            for(int j=0;j<n;j++)
            {
                if(board[i][j] == 'Y')
                    board[i][j] = 'O';
                else if(board[i][j] == 'O')
                    board[i][j] = 'X';
            }
    }
};