Surrounded Regions

最新推荐文章于 2022-06-01 17:41:20 发布

原创最新推荐文章于 2022-06-01 17:41:20 发布 · 781 阅读

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#深度 #广度

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本文详细介绍了使用深度优先和广度优先搜索算法解决棋盘问题的方法，包括基本思路、具体实现步骤及后期处理。通过实例演示，帮助读者理解如何通过搜索算法捕获被围死的棋子。

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);
	}