leetcode5/Surrounded_Regions.cpp at master · 1code/leetcode5 · GitHub

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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
*/

// The key observation is that the unsurrounded 'O' regions must have cells along
// one of the four edges of the square
// by Ke Hu (mrhuke@gmail.com) Apr. 2013
// O(n^2) time, O(n^2) space solution
class Solution {
public:

    void fill(vector<vector<char>> &board, int i, int j, queue<int> &q)
    {
        if (i >= 0 && i < board.size() && j >= 0 && j < board.size() && board[i][j] == 'O'){
            board[i][j] = '.';
            q.push(i*board.size() + j);
        }
    }

    void bfs(vector<vector<char>> &board, int i, int j, queue<int> &q)
    {
        fill(board, i, j, q);

        while(!q.empty()){
            int cur = q.front();
            q.pop();

            int r = cur / board.size(), c = cur % board.size();
            fill(board, r-1, c, q);
            fill(board, r+1, c, q);
            fill(board, r, c-1, q);
            fill(board, r, c+1, q);
        }
    }

    void solve(vector<vector<char>> &board) {
        if (board.empty()) return;

        queue<int> q;

        for( int i = 0; i < board.size(); ++i){
            bfs(board, 0, i, q);
            bfs(board, board.size()-1, i, q);
        }

        for( int i = 1; i < board.size()-1; ++i){
            bfs(board, i, 0, q);
            bfs(board, i, board.size()-1, q);
        }

        for (int i = 0; i < board.size(); ++i){
            for ( int j = 0; j < board.size(); ++j){
                if (board[i][j] == '.') board[i][j] = 'O';
                else if (board[i][j] == 'O') board[i][j] = 'X';
            }
        }
    }
};

// another solution based on DFS but cannot pass large test (mrhuke@gmail.com)
class Solution {
public:

    void fill(vector<vector<char>> &board, int i, int j, queue<int> &q)
    {
        if (i >= 0 && i < board.size() && j >= 0 && j < board.size() && board[i][j] == 'O'){
            board[i][j] = '.';
            q.push(i*board.size() + j);
        }
    }

    void dfs(vector<vector<char>> &board, int i, int j, set<int> &visited)
    {
        if (i >=0 && i < board.size() && j >= 0 && j < board.size() &&
        board[i][j] == 'O' && visited.find(i*board.size()+j) == visited.end())
        {
            board[i][j] = '.';
            visited.insert(i*board.size() + j);

            dfs(board, i-1, j, visited);
            dfs(board, i+1, j, visited);
            dfs(board, i, j-1, visited);
            dfs(board, i, j+1, visited);
        }
    }

    void solve(vector<vector<char>> &board) {
        if (board.empty()) return;

        set<int> visited;

        for( int i = 0; i < board.size(); ++i){
            dfs(board, 0, i, visited);
            dfs(board, board.size()-1, i, visited);
        }

        for( int i = 1; i < board.size()-1; ++i){
            dfs(board, i, 0, visited);
            dfs(board, i, board.size()-1, visited);
        }

        for (int i = 0; i < board.size(); ++i){
            for ( int j = 0; j < board.size(); ++j){
                if (board[i][j] == '.') board[i][j] = 'O';
                else if (board[i][j] == 'O') board[i][j] = 'X';
            }
        }
    }
};