华为OD机试E卷-机器人活动区域（100分） JAVA

解题思路：dfs搜索，笨办法。for循环二维数组，从每个点出发，把每个点能走到的距离记录到一个list里，最后对这个list进行求最大值即可。

import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;

public class OJTest11 {

    static int[][] directions = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};//移动的四个方向，上下左右

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int M = in.nextInt();
        int N = in.nextInt();
        int[][] maze = new int[M][N];
        for (int i = 0; i < maze.length; i++) {
            for (int j = 0; j < maze[0].length; j++) {
                maze[i][j] = in.nextInt();
            }
        }
        List<Integer> result = new ArrayList<>();
        for (int i = 0; i < maze.length; i++) {
            for (int j = 0; j < maze[0].length; j++) {
                int rs = solveMachine(maze, i, j);
                result.add(rs);
            }
        }
        System.out.println(result.stream().max(Integer::compareTo).orElse(null));
    }

    private static int solveMachine(int[][] maze, int i, int j) {
        //定义一个boolean的二维数组来记录路径是否走过了，走过了标记为true
        int M = maze.length;
        int N = maze[0].length;
        int count =0;
        boolean[][] visited = new boolean[M][N];
        if (!dfs(maze, i, j, visited)) {
            for (boolean[] bs : visited) {
                for (boolean b : bs) {
                    if (b) {
                        count++;
                    }
                }
            }
        }
        return count;
    }

    public static boolean dfs(int[][] maze, int i, int j, boolean[][] visited) {
        visited[i][j] = true; //进来函数先标记当前点
        //递归中止条件，发现上下左右都不能走就是中止条件
        for (int[] dir : directions) {
            int newX = i + dir[0];
            int newY = j + dir[1];
            if (newX >= 0 && newX < maze.length && newY >= 0 && newY < maze[0].length && !visited[newX][newY]) {
                if (Math.abs(maze[i][j] - maze[newX][newY]) <= 1) {
                    dfs(maze, newX, newY, visited);
                }
            }
        }
        //走到这说明四个方向都尝试了走不通，直接返回上层即可
        return false;
    }
}