题解：Surrounded Regions

最新推荐文章于 2021-03-11 10:12:34 发布

原创最新推荐文章于 2021-03-11 10:12:34 发布 · 331 阅读

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本文介绍一种算法，用于将二维棋盘中被X包围的所有O型区域翻转为X。通过先标记边缘及其相连的O，再将未被标记的O翻转成X来实现。

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题目如下：

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

解题思路上，先把边缘的O以及其附近的O标记出来，然后把被标记的O改回O的同时把未被标记的O（也就是被X包围的那些O）改成X，代码如下：

void turn(vector<vector<char> >&board, int i, int j) {
	if(board[i][j] == 'O'){
		board[i][j] = 'Y';
		if(i + 1 < board.size())
			turn(board, i + 1, j);
		if(j + 1 < board[0].size())
			turn(board, i, j + 1);
		if(i > 1)
			turn(board, i - 1, j);
		if(j > 1)
			turn(board, i, j - 1);
	}
}


void solve(vector<vector<char> >& board) {
	if (board.size() == 0)
	return;
    
    //row
    for (int i = 0; i < board[0].size(); i++) {
        if (board[0][i] == 'O')
            turn(board, 0, i);
            
        if (board[board.size() - 1][i] == 'O')
            turn(board, board.size() - 1, i);
    }
    
    //col
    for (int i = 0; i < board.size(); i++) {
        if (board[i][0] == 'O')
            turn(board, i, 0);
        if (board[i][board[0].size() - 1] == 'O')
            turn(board, i, board[0].size() - 1);
    }
    
    //last change
    for (int i = 0; i < board.size(); i++) {
        for (int j = 0; j < board[0].size(); j++) {
			if (board[i][j] == 'O') board[i][j] = 'X';
			if (board[i][j] == 'Y') board[i][j] = 'O';
    	}
	}
}