130. Surrounded Regions（BFS）

最新推荐文章于 2022-06-01 17:41:20 发布

Chihyung

最新推荐文章于 2022-06-01 17:41:20 发布

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分类专栏： Leetcode 文章标签： leetcode java

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本文介绍了一种解决二维棋盘上被包围的'O'字符问题的算法，通过使用宽度优先搜索(BFS)来标记边缘连接的'O'，避免它们被错误地转换为'X'。

题目：

Given a 2D board containing 'X' and 'O' (the letter 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

将被包围的‘O’变为‘X’

先将联通边框的‘O’变为‘Y’，然后将'O'变为‘X’，最后将‘Y’变为‘O’,采用BFS

代码：

class pair{
	int x,y;
	public pair(int x,int y)
	{
		this.x=x;
		this.y=y;
	}
}
public class Solution {
    public void solve(char[][] board) {
        if(board == null || board.length <=2 || board[0]==null || board[0].length <=2)return;
        LinkedList<pair> queue = new LinkedList<pair>();
        for(int i=0;i<board.length;i++)
        {
        	for(int j=0;j<board[0].length;j++)
        	{
        		if(i==0 || j==0 || i==(board.length-1) || j==(board[0].length-1))
        		{
        			if(board[i][j]=='O')queue.add(new pair(i, j));
        		}
        	}
        }
        
        while(!queue.isEmpty())
        {
        	pair cur = queue.pop();
        	int x=cur.x;
        	int y=cur.y;
        	if(board[x][y]=='Y')continue;
        	board[x][y]='Y';
        	if(y+1<board[0].length && board[x][y+1]=='O') queue.add(new pair(x, y+1));
        	if(y-1>=0 && board[x][y-1]=='O')queue.add(new pair(x, y-1));
        	if(x+1<board.length && board[x+1][y]=='O')queue.add(new pair(x+1, y));
        	if(x-1>=0 && board[x-1][y]=='O')queue.add(new pair(x-1, y));
        }
        
        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]=='Y')board[i][j]='O';
        	}
        }
    }
}