被围绕的区域

/*
  思路：
  使用Stream API来处理二维数组的操作。因为Stream API不直接支持二维数组操作，这里主要是通过传统遍历后转换为Stream来实现
*/
import java.util.stream.Stream;
import java.util.stream.IntStream;
 
public class SolutionStream {
    public void solve(char[][] board) {
        if (board == null || board.length == 0) return;
        int m = board.length;
        int n = board[0].length;
        // 标记边界上的'O'以及与其相连的'O'
        Stream.concat(
            IntStream.range(0, m).boxed(),
            IntStream.range(0, n).boxed()
        ).forEach(i -> {
            if (i < m && board[i][0] == 'O') dfs(board, i, 0);
            if (i < m && board[i][n - 1] == 'O') dfs(board, i, n - 1);
            if (i < n && board[0][i] == 'O') dfs(board, 0, i);
            if (i < n && board[m - 1][i] == 'O') dfs(board, m - 1, i);
        });
        // 将所有的'O'变为'X'
        IntStream.range(0, m).forEach(i ->
            IntStream.range(0, n).forEach(j -> {
                if (board[i][j] == 'O') board[i][j] = 'X';
            })
        );
        // 将所有的'*'还原为'O'
        IntStream.range(0, m).forEach(i ->
            IntStream.range(0, n).forEach(j -> {
                if (board[i][j] == '*') board[i][j] = 'O';
            })
        );
    }
    // 深度优先搜索，标记与边界'O'相连的'O'
    private void dfs(char[][] board, int i, int j) {
        if (i < 0 || i >= board.length || j < 0 || j >= board[0].length || board[i][j] != 'O') return;
        board[i][j] = '*';
        dfs(board, i + 1, j);
        dfs(board, i - 1, j);
        dfs(board, i, j + 1);
        dfs(board, i, j - 1);
    }
}