被围绕的区域

12 Oct 2024 in Algorithm Pageviews

leetcode!

给你一个 m x n 的矩阵 board ，由若干字符 'X' 和 'O' 组成，捕获 所有 被围绕的区域：

• 连接：一个单元格与水平或垂直方向上相邻的单元格连接。
• 区域：连接所有 'O' 的单元格来形成一个区域。
• 围绕：如果您可以用 'X' 单元格 连接这个区域，并且区域中没有任何单元格位于 board 边缘，则该区域被 'X' 单元格围绕。

通过将输入矩阵 board 中的所有 'O' 替换为 'X' 来 捕获被围绕的区域。

示例1：

输入

board = [["X","X","X","X"],["X","O","O","X"],["X","X","O","X"],["X","O","X","X"]]

输出

board = [["X","X","X","X"],["X","X","X","X"],["X","X","X","X"],["X","O","X","X"]]

解释：

在上图中，底部的区域没有被捕获，因为它在board的边缘并且不能被围绕。

示例2：

输入

board = [["X"]]

输出

[["X"]]

代码实现如下

class Solution {
public:
    void solve(vector<vector<char>>& board)
    {
        int n = board.size();
        int m = board[0].size();
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (i == 0 || i == n - 1 || j == 0 || j == m - 1) {
                    if (board[i][j] == 'O') {
                        dfs(board, i, j);
                    }
                }
            }
        }

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

    vector<int> dir{-1, 0, 1, 0, -1};
    void dfs(vector<vector<char>> &board, int i, int j)
    {
        int n = board.size();
        int m = board[0].size();
        if (i < 0 || i >= n || j < 0 || j >= m ||
            board[i][j] == 'X' || board[i][j] == '#') {
            return;
        }

        board[i][j] = '#';
        for (int k = 0; k < 4; k++) {
            dfs(board, i + dir[k], j + dir[k + 1]);
        }
    }
};