Description

Difficulty: 1.5/2.0 URL: https://oj.leetcode.com/problems/surrounded-regions/

Solution

We BFS from each boundary position with ‘O’. If a position with ‘O’ is visited in the BFS, it must be reachable from boundary position with ‘O’. Then we keep it to ‘O’, while set the rest to ‘X’.

class Solution {
	int dir[4][2] = { {0,1}, {0,-1}, {1,0}, {-1,0} };
	bool legal(const pair<int, int>& pos, const vector<vector<char>>&board) const{
		if (pos.first == -1 || pos.second == -1 || pos.first == board.size() || pos.second == board[0].size() || board[pos.first][pos.second] == 'X')
			return false;
		return true;
	}
public:
	void solve(vector<vector<char>> &board) {
		if (board.empty())
			return;
		vector<vector<bool>> visited(board.size(), vector<bool>(board[0].size(), false));

		for (int row = 0; row < board.size(); ++ row){
			BFS(row, 0, board, visited);
			BFS(row, board[0].size()-1, board, visited);
		}
		for (int col = 0; col < board[0].size(); ++ col){
			BFS(0, col, board, visited);
			BFS(board.size()-1, col, board, visited);
		}    	
		for (int row = 0; row < board.size(); ++ row){
			for (int col = 0; col < board[0].size(); ++ col){
				if (board[row][col] == 'O' && !visited[row][col])
					board[row][col] = 'X';    			
			}
		}
	}
	void BFS(int row, int col, vector<vector<char>>& board, vector<vector<bool>>& visited){
		pair<int,int> pos = pair<int,int>(row, col);
		if (visited[row][col] || board[row][col] == 'X')
			return;
		queue<pair<int,int>>bfs_q;
		bfs_q.push(pos);
		visited[row][col] = true;;
		while(!bfs_q.empty()){
			pos = bfs_q.front();
			bfs_q.pop();
			for (int i = 0; i < 4; ++ i){
				pair<int,int>new_pos(pos.first + dir[i][0], pos.second + dir[i][1]);
				if (!legal(new_pos, board) || visited[new_pos.first][new_pos.second])
					continue;
				bfs_q.push(new_pos);
				visited[new_pos.first][new_pos.second] = true;
			}
		}
	}
};

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