LeetCode（130） Surrounded Regions

Posted 2022-06-22 逆風的薔薇

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

分析

该题目的核心思想：只有在边界由‘O’连成的片区不会被改变为‘X’，其余位置的‘O’均会被改为‘X’；所以依次从每行的左右边界，每列的上下边界出发，检查元素，对于不能由‘O’改为‘X’的位置标记为'1'；

代码

class Solution 
public:
	void solve(vector<vector<char>>& board) 
		if (board.empty())
		
			return;
		

		int m = board.size(), n = board[0].size();
		/*核心思想，只有当边界节点为O*/
		for (int i = 0; i < m; ++i)
		
			//检查每行第一个元素
			check(board, i, 0, m, n);
			//若矩阵多列，则继续检查每行最后一个元素
			if (n > 1)
				check(board, i, n - 1, m, n);
		

		for (int j = 1; j < n-1; ++j)
		
			//检查每列第一个元素
			check(board, 0, j, m, n);

			//若矩阵多行，继续检查每列最底元素
			if (m > 1)
				check(board, m - 1, j, m, n);
		

		for (int i = 0; i < m; ++i)
		
			for (int j = 0; j < n; ++j)
			
				if (board[i][j] == 'O')
					board[i][j] = 'X';
			//for
		//for

		for (int i = 0; i < m; ++i)
		
			for (int j = 0; j < n; ++j)
			
				if (board[i][j] == '1')
					board[i][j] = 'O';
			//for
		//for
		
	

	/*检查（i,j）位置是否为'o',若为'o'，标记为'1' */
	void check(vector<vector<char>>& board,int i, int j, int row, int col)
	
		if (i < 0 || i >= row || j < 0 || j >= col)
			return;

		if (board[i][j] == 'O')
		
			board[i][j] = '1';
			if (i > 1)
				check(board, i - 1, j, row, col);
			
			if (i < row - 1)
				check(board, i + 1, j, row, col);

			if (j > 1)
				check(board, i, j - 1, row, col);

			if (j < col - 1)
				check(board, i, j + 1, row, col);
		
	
;

GitHub源码



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