Leetcode No.130 被围绕的区域（DFS）

输入：board = [["X","X","X","X"],["X","O","O","X"],["X","X","O","X"],["X","O","X","X"]]
 输出：[["X","X","X","X"],["X","X","X","X"],["X","X","X","X"],["X","O","X","X"]]
 解释：被围绕的区间不会存在于边界上，换句话说，任何边界上的 'O' 都不会被填充为 'X'。 任何不在边界上，或不与边界上的 'O' 相连的 'O' 最终都会被填充为 'X'。如果两个元素在水平或垂直方向相邻，则称它们是“相连”的。
示例 2：
 输入：board = [["X"]]
 输出：[["X"]]
提示：
m == board.length
 n == board[i].length
 1 <= m, n <= 200
 board[i][j] 为 'X' 或 'O'

X X X X
X O O X
X X X X
X O O X

X X X X
X # # X
X X X X
X O O X

public class Solution {
    int n, m;
    public void solve(char[][] board) {
        n = board.length;//行
        if (n == 0) {
            return;
        }
        m = board[0].length;//列
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                // 从边缘o开始搜索
                boolean isEdge = (i == 0 || j == 0 || i == n-1 || j == m-1);
                if (isEdge && board[i][j] == 'O') {
                    dfs(board, i, j);
                }
            }
        }
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                //board[i][j] == '#' 边界上的O,或者与之连通的O,不需要改变
                if (board[i][j] == '#') {
                    board[i][j] = 'O';
                } else if (board[i][j] == 'O') {
                    board[i][j] = 'X';
                }
            }
        }
    }
    //寻找边界上的O,或者与之连通的O
    public void dfs(char[][] board, int x, int y) {
        if (x < 0 || x >= n || y < 0 || y >= m || board[x][y] != 'O') {
            return;
        }
        board[x][y] = '#';//边界上的O与与之连通的0特殊处理
        dfs(board, x + 1, y);//下
        dfs(board, x - 1, y);//上
        dfs(board, x, y + 1);//右
        dfs(board, x, y - 1);//左
    }
}