0490. The Maze

https://leetcode.com/problems/the-maze

Description

There is a ball in a maze with empty spaces (represented as 0) and walls (represented as 1). The ball can go through the empty spaces by rolling up, down, left or right, but it won't stop rolling until hitting a wall. When the ball stops, it could choose the next direction.

Given the m x n maze, the ball's start position and the destination, where start = [startrow, startcol] and destination = [destinationrow, destinationcol], return true if the ball can stop at the destination, otherwise return false.

You may assume that the borders of the maze are all walls (see examples).

Example 1:

**Input:** maze = [[0,0,1,0,0],[0,0,0,0,0],[0,0,0,1,0],[1,1,0,1,1],[0,0,0,0,0]], start = [0,4], destination = [4,4]
**Output:** true
**Explanation:** One possible way is : left -> down -> left -> down -> right -> down -> right.

Example 2:

**Input:** maze = [[0,0,1,0,0],[0,0,0,0,0],[0,0,0,1,0],[1,1,0,1,1],[0,0,0,0,0]], start = [0,4], destination = [3,2]
**Output:** false
**Explanation:** There is no way for the ball to stop at the destination. Notice that you can pass through the destination but you cannot stop there.

Example 3:

**Input:** maze = [[0,0,0,0,0],[1,1,0,0,1],[0,0,0,0,0],[0,1,0,0,1],[0,1,0,0,0]], start = [4,3], destination = [0,1]
**Output:** false

Constraints:

• m == maze.length

• n == maze[i].length

• 1 <= m, n <= 100

• maze[i][j] is 0 or 1.

• start.length == 2

• destination.length == 2

• 0 <= startrow, destinationrow <= m

• 0 <= startcol, destinationcol <= n

• Both the ball and the destination exist in an empty space, and they will not be in the same position initially.

• The maze contains at least 2 empty spaces.

ac

class Solution {
    public boolean hasPath(int[][] maze, int[] start, int[] destination) {
        int row = maze.length, col = maze[0].length;
        boolean[][] visited = new boolean[row][col];
        return dfs(maze, start[0], start[1], destination, visited);
    }

    private boolean dfs(int[][] maze, int r, int c, int[] dest, boolean[][] visited) {
        // exit, reach destination

        visited[r][c] = true;
        // 4 directions
        for (int i = 0; i < 4; i++) {
            int[] loc = rolling(maze, r, c, i);
            if (loc[0] == dest[0] && loc[1] == dest[1]) return true; // hit the end
            if (visited[loc[0]][loc[1]]) continue; // skip visitd cell
            if (dfs(maze, loc[0], loc[1], dest, visited)) return true; // find one valid path
        }

        return false; // can't find path
    }

    private int[] rolling(int[][] maze, int r, int c, int direction) {
        // 0: up, 1: down, 2:left, 3:right
        if (direction == 0) {
            while (r - 1 >= 0 && maze[r-1][c] == 0) r--;
        } else if (direction == 1) {
            while (r + 1 < maze.length && maze[r+1][c] == 0) r++;
        } else if (direction == 2) {
            while (c - 1 >= 0 && maze[r][c-1] == 0) c--;
        } else if (direction == 3) {
            while (c + 1 < maze[0].length && maze[r][c+1] == 0) c++;
        }

        return new int[]{r, c};
    }
}

/*
just dfs 4 directions, with helper rolling() func
*/