Surrounded Regions

题目原型：

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

基本思路：

这个题的意思就是类似于，给你一张围棋图，如果‘O’表示白子，'X'表示黑子，问，哪些白子被围死了。自然我们想到了深度优先和广度优先搜索，那么从哪里开始搜索了，当然是从棋盘的四周了，搜索完后把被围死的白子变为黑子即可。

关于深度和广度优先搜索算法网上有很多资料，大体上可以分为两种，一种是递归另一种是非递归，在这里我们用的是非递归做法。

	//广度优先搜索
	public void bfs(char[][] board, int low , int col)
	{
		int lenOfcol = board[0].length;
		Queue<Integer> queue = new LinkedList<Integer>();
		visit(board, low, col, queue);
		while(!queue.isEmpty())
		{
			int current = queue.poll();
			int i = current/lenOfcol;
			int j = current%lenOfcol;
			
			visit(board, i-1, j, queue);
			visit(board, i, j-1, queue);
			visit(board, i+1, j, queue);
			visit(board, i, j+1, queue);
			
		}
	}
	//深度优先搜索
	public void dfs(char[][] board, int low , int col)
	{
		Stack<Integer> stack = new Stack<Integer>();
		int lenOflow = board.length;
		int lenOfcol = board[0].length;
		if(!stack.contains(low*lenOfcol+col))
		{
			visit(board, low, col, stack);
		}
		while(!stack.isEmpty())
		{
			int current = stack.peek();
			int i = current/lenOfcol;
			int j = current%lenOfcol;
			if(i-1>=0&&i-1<lenOflow&&board[i-1][j]=='O')
			{
				visit(board, i-1, j, stack);
				continue;
			}
			if(j-1>=0&&j-1<lenOfcol&&board[i][j-1]=='O')
			{
				visit(board, i, j-1, stack);
				continue;
			}
			if(i+1>=0&&i+1<lenOflow&&board[i+1][j]=='O')
			{
				visit(board, i+1, j, stack);
				continue;
			}
			if(j+1>=0&&j+1<lenOfcol&&board[i][j+1]=='O')
			{
				visit(board, i, j+1, stack);
				continue;
			}
			else
				stack.pop();
		}
	}

	public void visit(char[][] board, int low , int col,Stack<Integer> stack)
	{
		int lenOflow = board.length;
		int lenOfcol = board[0].length;
		
		if(low<0||low>=lenOflow||col<0||col>=lenOfcol||board[low][col]!='O')
			return;
		board[low][col] = '+';
		stack.push(low*lenOfcol+col);
	}

	public void visit(char[][] board, int low , int col,Queue<Integer> queue)
	{
		int lenOflow = board.length;
		int lenOfcol = board[0].length;
		
		if(low<0||low>=lenOflow||col<0||col>=lenOfcol||board[low][col]!='O')
			return;
		board[low][col] = '+';
		queue.offer(low*lenOfcol+col);
	}
	
	//后期处理
	public void doit(char[][] board)
	{
		for(int i = 0;i<board.length;i++)
		{
			for(int j = 0;j<board[0].length;j++)
			{
				if(board[i][j]=='O')
					board[i][j]='X';
				if(board[i][j]=='+')
					board[i][j]='O';
			}
		}
	}

	public void solve(char[][] board)
	{
		if(board==null||board.length==0||board[0].length==0)
			return;
		int lenOflow = board.length;//行长
		int lenOfcol = board[0].length;//列长
		
		//第一行，第一列，第lenOflow行，第lenOfcol列
		for(int col = 0;col<lenOfcol;col++)
		{
			/*bfs(board, 0, col);
			bfs(board, lenOflow-1, col);*/
			dfs(board, 0, col);
			dfs(board, lenOflow-1, col);
		}
		for(int low = 1;low<lenOflow-1;low++)//此处要特别注意low的起点和终点
		{
			/*bfs(board, low, 0);
			bfs(board, low, lenOfcol-1);*/
			dfs(board, low, 0);
			dfs(board, low, lenOfcol-1);
		}
		doit(board);
	}