Leetcode Surrounded Regions

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

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
  
   > &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
   
     > &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
    
      > &board, int i, int j) { return !(i<0 || i>=board.size() || j<0 || j>=board[0].size() || board[i][j] != 'O'); }