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 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

思路：引入一个新的符号，比如“*”。先把四周的圈都保存下来。然后再填充和变换

public class Solution {
    public void solve(char[][] board) {
        if (board.length < 2 || board[0].length < 2)
            return;
        int m = board.length, n = board[0].length;
        //Any 'O' connected to a boundary can't be turned to 'X', so ...
        //Start from first and last column, turn 'O' to '*'.
        for (int i = 0; i < m; i++) {
            if (board[i][0] == 'O')
                boundaryDFS(board, i, 0);
            if (board[i][n-1] == 'O')
                boundaryDFS(board, i, n-1);    
        }
        //Start from first and last row, turn '0' to '*'
        for (int j = 0; j < n; j++) {
            if (board[0][j] == 'O')
                boundaryDFS(board, 0, j);
            if (board[m-1][j] == 'O')
                boundaryDFS(board, m-1, j);    
        }
        //post-prcessing, turn 'O' to 'X', '*' back to 'O', keep 'X' intact.
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (board[i][j] == 'O')
                    board[i][j] = 'X';
                else if (board[i][j] == '*')
                    board[i][j] = 'O';
            }
        }
    }
    //Use DFS algo to turn internal however boundary-connected 'O' to '*';
    private void boundaryDFS(char[][] board, int i, int j) {
        if (i < 0 || i > board.length - 1 || j <0 || j > board[0].length - 1)
            return;
        if (board[i][j] == 'O')
            board[i][j] = '*';
        if (i > 1 && board[i-1][j] == 'O')
            boundaryDFS(board, i-1, j);
        if (i < board.length - 2 && board[i+1][j] == 'O')
            boundaryDFS(board, i+1, j);
        if (j > 1 && board[i][j-1] == 'O')
            boundaryDFS(board, i, j-1);
        if (j < board[i].length - 2 && board[i][j+1] == 'O' )
            boundaryDFS(board, i, j+1);
    }
}