124_leetcode_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 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

1:注意特殊情況；2:將board四邊上的0， 以及與這些‘0’相鄰的‘0’都設置爲‘C'；2:把board中的其他的‘0’（這些‘0’全部被‘X'包圍）設置爲‘X'; 3:把board中的‘C'全部設置爲‘0’

          void solve(vector<vector<char>> &board) {
        if(board.size() == 0 || board.size() == 1 || board[0].size() == 1 || board.size() == 2 || board[0].size() == 2)
            return;
        
        int rows = (int)board.size();
        int columns = (int)board[0].size();
        
        for(int i = 0; i < columns; i++)
        {
            if(board[0][i] == 'O')
                findUnchangedO(board, 0, i);
        }
        
        for(int i = 0; i < rows; i++)
        {
            if(board[i][0] == 'O')
                findUnchangedO(board, i, 0);
        }
        
        for(int i = 0; i < columns; i++)
        {
            if(board[rows-1][i] == 'O')
                findUnchangedO(board, rows-1, i);
        }
        
        for(int i = 0; i < rows; i++)
        {
            if(board[i][columns-1] == 'O')
                findUnchangedO(board, i, columns-1);
        }
        
        for(int i = 0; i < rows; i++)
        {
            for(int j = 0; j < columns; j++)
                if(board[i][j] == 'O')
                    board[i][j] = 'X';
        }
        
        for(int i = 0; i < rows; i++)
        {
            for(int j = 0; j < columns; j++)
                if(board[i][j] == 'C')
                    board[i][j] = 'O';
        }
        
        return ;
    }
    
    void findUnchangedO(vector<vector<char> > &board, int curX, int curY)
    {
        board[curX][curY] = 'C';
        
        if(curX-1 > 0 && board[curX-1][curY] == 'O')
            findUnchangedO(board, curX-1, curY);
        if(curY - 1 > 0 && board[curX][curY - 1] == 'O')
            findUnchangedO(board, curX, curY - 1);
        
        if(curX + 1 < board.size() && board[curX + 1][curY] == 'O')
            findUnchangedO(board, curX + 1, curY);
        if(curY + 1 < board[0].size() && board[curX][curY + 1] == 'O')
            findUnchangedO(board, curX, curY + 1);
    }