130. Surrounded Regions

Given a 2D board containing ‘X’ and ‘O’ (the letter 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 XX O O XX X O XX O X X

After running your function, the board should be:

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

class Solution {    public void solve(char[][] board) {        if(board.length <= 2 || board[0].length <= 2 )             return;        int row = board.length;        int column = board[0].length;        for(int i = 0; i < row; i++){            if(board[i][0] == 'O'){                boundarySearch(board, i , 0);            }            if(board[i][column - 1] == 'O'){                boundarySearch(board, i, column-1);            }        }        for(int j = 0; j < column; j++){            if(board[0][j] == 'O'){                boundarySearch(board, 0, j);            }            if(board[row-1][j] == 'O'){                boundarySearch(board, row - 1, j);            }        }        for(int i = 0; i < row; i++){            for(int j = 0; j < column; j++){                if(board[i][j] == 'O'){                    board[i][j] = 'X';                }else if(board[i][j] == '*'){                    board[i][j] = 'O';                }            }        }    }    public void boundarySearch(char[][] board, int i, int j){        if(i < 0 || j < 0 || i >= board.length || j >= board[0].length)             return;        if(board[i][j] == 'X' || board[i][j] == '*')             return;        board[i][j] = '*';        if(i > 1 && board[i - 1][j] == 'O')             boundarySearch(board, i - 1, j);        if(j > 1 && board[i][j - 1] == 'O')             boundarySearch(board, i, j - 1);        if(i < board.length-2 && board[i + 1][j] == 'O')             boundarySearch(board, i + 1, j);        if(j < board[0].length-2 && board[i][j + 1] == 'O')             boundarySearch(board, i, j + 1);    }}