Surrounded Regions - LeetCode

目錄

• 題目連接
• 注意點
• 解法
• 小結

題目連接

https://leetcode.com/problems/surrounded-regions/

注意點

• 邊緣不算包圍‘O’

解法

解法一：dfs。找處在邊緣上的O而後dfs將與之相鄰的O都改成#。處理完以後再把這時候的O改成X,#改成O便可

class Solution {
public:
    void solve(vector<vector<char>>& board) {
        int n  = board.size();
        for(int i = 0;i < n;i++)
        {
            for(int j = 0;j < board[i].size();j++)
            {
                if((i == 0 || i == n-1 || j == 0 ||j == board[i].size()-1) && board[i][j] == 'O') dfs(board,i,j);
            }
        }
        for(int i = 0;i < n;i++)
        {
            for(int j = 0;j < board[i].size();j++)
            {
                if(board[i][j] == 'O') board[i][j] = 'X';
                if(board[i][j] == '#') board[i][j] = 'O';
            }
        }
    }
    void dfs(vector<vector<char>>& board,int i,int j)
    {
        if (board[i][j] == 'O')
        {
            board[i][j] = '#';
            if (i > 0 && board[i - 1][j] == 'O') 
                dfs(board, i - 1, j);
            if (j < board[i].size() - 1 && board[i][j + 1] == 'O') 
                dfs(board, i, j + 1);
            if (i < board.size() - 1 && board[i + 1][j] == 'O') 
                dfs(board, i + 1, j);
            if (j > 0 && board[i][j - 1] == 'O') 
                dfs(board, i, j - 1);
        }
    }
};

解法二：bfs。基本上同樣的思路，仍是找處在邊緣上的O而後bfs將與之相鄰的O都改成#。處理完以後再把這時候的O改成X,#改成O便可

class Solution {
public:
    void solve(vector<vector<char>>& board) {
        int n  = board.size();
        for(int i = 0;i < n;i++)
        {
            for(int j = 0;j < board[i].size();j++)
            {
                if((i == 0 || i == n-1 || j == 0 ||j == board[i].size()-1) && board[i][j] == 'O') dfs(board,i,j);
            }
        }
        for(int i = 0;i < n;i++)
        {
            for(int j = 0;j < board[i].size();j++)
            {
                if(board[i][j] == 'O') board[i][j] = 'X';
                if(board[i][j] == '#') board[i][j] = 'O';
            }
        }
    }
    void dfs(vector<vector<char>>& board,int i,int j)
    {
        if (board[i][j] == 'O')
        {
            board[i][j] = '#';
            if (i > 0 && board[i - 1][j] == 'O') 
                dfs(board, i - 1, j);
            if (j < board[i].size() - 1 && board[i][j + 1] == 'O') 
                dfs(board, i, j + 1);
            if (i < board.size() - 1 && board[i + 1][j] == 'O') 
                dfs(board, i + 1, j);
            if (j > 0 && board[i][j - 1] == 'O') 
                dfs(board, i, j - 1);
        }
    }
};

小結

• 這道題dfs和bfs效率都差很少