Leetcode Surrounded Regions

解决包围的棋盘问题

最新推荐文章于 2018-05-07 18:36:05 发布

iteye_18800

最新推荐文章于 2018-05-07 18:36:05 发布

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本文探讨了如何解决二维棋盘上被X包围的O区域的捕获问题，通过四面搜索算法，实现O区域的识别与转换。关键在于理解算法逻辑与优化搜索路径。

Surrounded Regions

Total Accepted:4258 Total Submissions:30340 My Submissions

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

本题关键是理解问题的特点：四面的最外层搜索，有O通路的就为不被包围的区域，其他都可以置为X；

没把握这个特点，难度为五星级的。

知道这个特点，难度瞬间变为3到4星级。

//2014-2-18 update
	const static char NON_SURROUNDED = '*';
	void solve(vector<vector<char>> &board) 
	{
		if (board.empty() || board[0].empty()) return;

		for (int i = 0; i < board.size(); i++)
		{
			backtrack(board, i, 0);
			backtrack(board, i, board[0].size()-1);
		}
		for (int i = 1; i < board[0].size(); i++)
		{
			backtrack(board, 0, i);
			backtrack(board, board.size()-1, i);
		}
		for (int i = 0; i < board.size(); i++)
		{
			for (int j = 0; j < board[0].size(); j++)
			{
				if (board[i][j] == NON_SURROUNDED)  board[i][j] = 'O';
				else board[i][j] = 'X';
			}
		}

	}
	void backtrack(vector<vector<char> > &board, int row, int col)
	{
		if (!isLegal(board, row, col)) return;
		board[row][col] = NON_SURROUNDED;
		backtrack(board, row+1, col);
		backtrack(board, row-1, col);
		backtrack(board, row, col+1);
		backtrack(board, row, col-1);
	}
	bool isLegal(vector<vector<char> > &board, int i, int j)
	{
		return 
		!(i<0 || i>=board.size() || j<0 || j>=board[0].size() || board[i][j] != 'O');
	}