【leetcode】Surrounded Regions

鏈接：https://oj.leetcode.com/problems/surrounded-regions/

描述：

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

解法一：

本體開始一直在想着怎麼找到所有的包圍O的環，想了很久沒想出來。

後面從分析怎樣的O不需要flip，就有了解法。

第一個是dfs遞歸的，提交是Runtime error，應該是棧溢出的問題。

代碼如下：

    void dfs(vector<vector<char>> &board, int i, int j)
    {
    	int m = board.size();
    	int n = board.at(0).size();
    	if( i < 0 || i >= m ||
    		j < 0 || j >= n ||
    		board[i][j] != 'O')
    		return;
    	board[i][j] = 'E';
    	dfs(board, i-1, j);
    	dfs(board, i+1, j);
    	dfs(board, i,  j-1);
    	dfs(board, i, j + 1);
    }
    
    void solve(vector<vector<char>> &board)
    {
    	int m= board.size();
    	if( m <= 0) return;
    	int n = board.at(0).size();
    	for(int i= 0; i < m; ++i)
    	{
    		dfs(board, i, 0);
    		dfs(board, i, n - 1);
    	}
    
    	for(int i=0; i < n; ++i)
    	{
    		dfs(board, 0, i);
    		dfs(board, m-1, i);
    	}
    	for(int i=0; i < m; ++i)
    	{
    		for(int j = 0; j < n; ++j)
    		{
    			if( board[i][j] == 'O')
    				board[i][j] = 'X';
    			else if(board[i][j] == 'E')
    				board[i][j] = 'O';
    		}
    	}
    }

方法二：

非遞歸版的，用stack實現，用到一個技巧：怎麼用一個數存儲座標的i和j兩個值。

代碼如下：

void dfs2(vector<vector<char>> &board, int i, int j)
    {
    	if( board[i][j] != 'O' )
    		return;
    	int m= board.size();
    	int n =  board[0].size();
    	int big = max(m, n);
    	stack<int> q;
    	q.push(i * big + j);
    
    	int next[4][2] = { -1, 0,	1, 0, 0, -1, 0, 1};
    	while(!q.empty())
    	{
    		int num = q.top();
    		q.pop();
    		int curI = num/big;
    		int curJ = num%big;
    		board[curI][curJ] = 'E';
    		for(int count=0; count < 4; ++count)
    		{
    			int nextI = curI + next[count][0];
    			int nextJ = curJ + next[count][1];
    			if( nextI < 0 || nextI >= m || 	nextJ < 0 || nextJ >= n || 
    				board[nextI][nextJ] != 'O')
    				continue;
    				q.push(nextI * big + nextJ);
    		}
    	}
    }
    
    void solve(vector<vector<char>> &board)
    {
    	int m= board.size();
    	if( m <= 0) return;
    	int n = board.at(0).size();
    	for(int i= 0; i < m; ++i)
    	{
    		dfs2(board, i, 0);
    		dfs2(board, i, n - 1);
    	}
    
    	for(int i=0; i < n; ++i)
    	{
    		dfs2(board, 0, i);
    		dfs2(board, m-1, i);
    	}
    	for(int i=0; i < m; ++i)
    	{
    		for(int j = 0; j < n; ++j)
    		{
    			if( board[i][j] == 'O')
    				board[i][j] = 'X';
    			else if(board[i][j] == 'E')
    				board[i][j] = 'O';
    		}
    	}
    }