Leetcode -- surrounded regions

Given a 2D Board containing‘X‘And‘O‘, Capture all regions surrounded‘X‘.

A region is captured by flipping all‘O‘S‘X‘S in that surrounded region.

For example,

X X X XX O O XX X O XX O X X

 

After running your function, the Board shoshould be:

X X X XX X X XX X X XX O X X

This is a simple problem, we just need to apply bfs on ‘O‘ elements on the edges of board. 

public class Solution {    public void solve(char[][] board) {        int row = board.length;if(row > 0){int column = board[0].length;Set<Integer> alreadyChecked = new HashSet<Integer>();for(int i = 0; i < column; ++i){if(board[0][i] == ‘O‘ && !alreadyChecked.contains(i))bfs(board,0, i, alreadyChecked);if(board[row - 1][i] == ‘O‘ && !alreadyChecked.contains((row - 1) * column + i))bfs(board,row - 1, i, alreadyChecked);}for(int i = 1; i < row - 1; ++i){if(board[i][0] == ‘O‘ && !alreadyChecked.contains(i * column))bfs(board,i, 0, alreadyChecked);if(board[i][column - 1] == ‘O‘ && !alreadyChecked.contains(i * column + column - 1))bfs(board,i, column - 1, alreadyChecked);}for(int i = 0; i < row; ++i){for(int j = 0; j < column; ++j){if(board[i][j] == ‘O‘ && !alreadyChecked.contains(i*column + j))board[i][j] = ‘X‘;}}}    }private void bfs(char[][] board, int i, int j, Set<Integer> alreadyChecked) {int row = board.length;if(row > 0){int column = board[0].length;if(!alreadyChecked.contains(i * column + j)) {Queue<Integer> current = new LinkedList<Integer>();alreadyChecked.add(i * column + j);current.add(i * column + j);while(current.peek() != null){int currentCoordinate = current.poll();int x = currentCoordinate / column;int y = currentCoordinate % column;if(x - 1 >= 0 && board[x - 1][y] == ‘O‘ && !alreadyChecked.contains(currentCoordinate - column)){current.add(currentCoordinate - column);alreadyChecked.add(currentCoordinate - column);}if(x + 1 < row && board[x + 1][y] == ‘O‘ && !alreadyChecked.contains(currentCoordinate + column)){current.add(currentCoordinate + column);alreadyChecked.add(currentCoordinate + column);}if(y - 1 >= 0 && board[x][y - 1] == ‘O‘ && !alreadyChecked.contains(currentCoordinate - 1)) {current.add(currentCoordinate - 1);alreadyChecked.add(currentCoordinate - 1);}if(y + 1 < column && board[x][y + 1] == ‘O‘ && !alreadyChecked.contains(currentCoordinate + 1)) {current.add(currentCoordinate + 1);alreadyChecked.add(currentCoordinate + 1);}}}}        }}