Leetcode Surrounded Regions

解决包围的区域问题

最新推荐文章于 2025-07-29 14:24:32 发布

最新推荐文章于 2025-07-29 14:24:32 发布 · 122 阅读

本文详细阐述了如何解决二维板中被X包围的O区域问题，通过深入理解问题特性和实现回溯算法来高效地捕获所有未被包围的区域，并进行区域翻转。文章从算法设计角度出发，提供了清晰的代码实现思路和步骤，旨在帮助读者掌握解决类似问题的方法。

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');
	}