/**
* https://oj.leetcode.com/problems/surrounded-regions/
* 从四个边界的'O'出发，它们所能到达的'O'就是没有被包围的'O'。
* 所以，该题能够用BFS遍历或者DFS遍历。
*/
class Solution {
public:
void solve(vector<vector<char>> &board) {
int row = board.size();
if(row <= 2){
return;
}
int col = board[0].size();
if(col <=2){
return;
}
int i = 0;
int j = 0;
if(row > 2 && col > 2){
for(i = 0; i < row; i++){
if(board[i][0] == 'O'){
board[i][0] = 'P';
if(board[i][1] == 'O'){
search(i, 1, row, col, board);
}
}
if(board[i][col - 1] == 'O'){
board[i][col - 1] = 'P';
if(board[i][col - 2] == 'O'){
search(i, col - 2, row, col, board);
}
}
}//end of for(i = 0; i < row; i++)
for(j = 0; j < col; j++){
if(board[0][j] == 'O'){
board[0][j] = 'P';
if(board[1][j] == 'O'){
search(1, j, row, col, board);
}
}
if(board[row - 1][j] == 'O'){
board[row - 1][j] = 'P';
if(board[row - 2][j] == 'O'){
search(row - 2, j, row, col, board);
}
}
} //end of for(j = 0; j < col; j++)
}//end of if(row >2 && col > 2)
for(i = 0; i < row; i++){
for(j = 0; j < col; j++){
if(board[i][j] == 'P'){
board[i][j] = 'O';
} else {
board[i][j] = 'X';
}
}
}
}
//search(1, j, row, col, board);
void search(int i, int j, const int row, const int col, vector<vector<char>> &board){
board[i][j]='P';
if(i - 1 > 0 && board[i - 1][j] == 'O'){
search(i - 1, j, row, col, board);
}
if(i + 1 < row - 1 && board[i + 1][j] == 'O'){
search(i+1, j, row, col, board);
}
if(j - 1 > 0 && board[i][j - 1] == 'O'){
search(i, j - 1, row, col, board);
}
if(j + 1 < col && board[i][j + 1] == 'O'){
search(i, j + 1, row, col, board);
}
}
};