Surrounded Regions

最新推荐文章于 2022-06-01 17:41:20 发布

zephyr_be_brave

最新推荐文章于 2022-06-01 17:41:20 发布

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分类专栏： leetcode习题分析文章标签： DFS BFS

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本文介绍了一种解决二维棋盘中'O'被'X'包围问题的方法，通过深度优先搜索（DFS）和广度优先搜索（BFS）两种算法实现，将所有被'X'包围的'O'翻转为'X'。

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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

思路很简单，就是从外圈向内圈搜索。

方法一：DFS，代码如下：

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

56 milli secs.

方法二：BFS，用队列实现

class Solution {
public:
    struct pos{
        int i;
        int j;
        public :pos(int x,int y)
        {
            i = x;
            j = y;
        }
    };
    void solve(vector<vector<char>> &board) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int m = board.size();
        if(m == 0) return;
        int n = board[0].size();
        queue<pos> queue;
        
        for(int i = 0;i < n;i++)
        {
            if(board[0][i] == 'O') queue.push(pos(0,i));
            if(board[m-1][i] == 'O') queue.push(pos(m-1,i));
        }
        for(int i = 1;i < m-1;i++)
        {
            if(board[i][0] == 'O') queue.push(pos(i,0));
            if(board[i][n-1] == 'O') queue.push(pos(i,n-1));
        }
        
        bfs(queue,board);
        for(int i = 0;i < m;i++)
        {
            for(int j = 0;j < n;j++)
            {
                if(board[i][j] == 'O') board[i][j] = 'X';
                else if(board[i][j] == 'S') board[i][j] = 'O';
            }
        }
    }
    
    void bfs(queue<pos>& queue,vector<vector<char>> &board)
    {
        while(!queue.empty())
        {
            pos cur = queue.front();
            queue.pop();
            board[cur.i][cur.j] = 'S';
            search(cur.i+1,cur.j,board,queue);
            search(cur.i-1,cur.j,board,queue);
            search(cur.i,cur.j+1,board,queue);
            search(cur.i,cur.j-1,board,queue);
        }
    }
    
    void search(int i,int j,vector<vector<char>> &board,queue<pos>& queue)
    {
         if(i < 0 || i >= board.size() || j < 0 || j >= board[0].size()
        ||board[i][j] != 'O') return;
        queue.push(pos(i,j));   
    }
};

68 milli secs.