/*** https://oj.leetcode.com/problems/surrounded-regions/* 從四個邊界的'O'出發，它們所能到達的'O'就是沒有被包圍的'O'。* 所以，該題能夠用BFS遍歷或者DFS遍歷。 */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);}} };

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