Surrounded Regions

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 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

Have you been asked this question in an interview?
这道题目的思路：
从上下左右四个边界找为O 的位置，从该位置往board 里面继续寻找相连的为O的位置，一直到走不通。把这些位置用“p”标记。四条边走一遍以后，board 上剩下的为'O'的位置 就是被'X'包围的位置。把剩余的'O'换成'X', 把'P' 换回原来的'O'.

实现往里走搜索的方式 可以用DSF, BFS 都可以。下面写法是用DFS, 如果用 bfs的话，需要用到queue。
需要注意的corner case ： 当行或者列小于等于2的时候。

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