Leetcode: Surrounded Regions

最新推荐文章于 2019-02-26 13:31:05 发布

原创最新推荐文章于 2019-02-26 13:31:05 发布 · 717 阅读

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本文介绍了一种解决二维棋盘中被'X'包围的'O'区域捕获问题的算法实现。通过递归地将边界上的'O'标记为临时字符，避免这些'O'被错误地转化为'X'，最终将所有未被标记的'O'转化为'X'，再将临时标记还原为'O'。

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

public class Solution {
    public void solve(char[][] board) {
		// Start typing your Java solution below
		// DO NOT write main() function
		if(board.length < 1)
			return;
		//change 'O' to 'A'
		//left -> right
		for(int j = 0; j < board[0].length; j++)
			flip(board, 0, j);
		//up -> down
		for(int i = 0; i < board.length; i++)
			flip(board, i, board[0].length - 1);
		//right -> left
		for(int j = board[0].length - 1; j >= 0; j--)
			flip(board, board.length - 1, j);
		//down -> up
		for(int i = board.length - 1; i >= 0; i--)
			flip(board, i, 0);
		//change 'O' to 'X'
		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';
		//change 'A' to 'O'
		for(int i = 0; i < board.length; i++)
			for(int j = 0; j < board[0].length; j++)
				if(board[i][j] == 'A')
					board[i][j] = 'O';
	}
	public void flip(char[][] board, int i, int j){
		if(i < 0 || i >= board.length || j < 0 || j >= board[0].length)
			return;
		if(board[i][j] == 'O'){
			board[i][j] = 'A';
			flip(board, i - 1, j);
			flip(board, i + 1, j);
			flip(board, i, j - 1);
			flip(board, i, j + 1);
		}
	}
}