每日一题-leetcode 被围绕的区域

给你一个 m x n 的矩阵 board ，由若干字符 'X' 和 'O' ，找到所有被 'X' 围绕的区域，并将这些区域里所有的 'O' 用 'X' 填充。

示例 1：

image.png

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'

java代码：

class Solution {
    int n, m;

    //深度优先搜索
    public void solve(char[][] board) {
        n = board.length;
        if (n == 0) {
            return;
        }
        m = board[0].length;
        for (int i = 0; i < n; i++) {
            dfs(board, i, 0);
            dfs(board, i, m - 1);
        }
        for (int i = 1; i < m - 1; i++) {
            dfs(board, 0, i);
            dfs(board, n - 1, i);
        }
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (board[i][j] == 'A') {
                    board[i][j] = 'O';
                } else if (board[i][j] == 'O') {
                    board[i][j] = 'X';
                }
            }
        }
    }

    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] = 'A';
        dfs(board, x + 1, y);
        dfs(board, x - 1, y);
        dfs(board, x, y + 1);
        dfs(board, x, y - 1);
    }
}