[LeetCode] 490. The Maze_Medium tag: BFS/DFS

There is a ball in a maze with empty spaces and walls. The ball can go through 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 ball's start position, the destination and the maze, determine whether the ball could stop at the destination.

The maze is represented by a binary 2D array. 1 means the wall and 0 means the empty space. You may assume that the borders of the maze are all walls. The start and destination coordinates are represented by row and column indexes.

Example 1

Input 1: a maze represented by a 2D array
 
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
 
Input 2: start coordinate (rowStart, colStart) = (0, 4)
Input 3: destination coordinate (rowDest, colDest) = (4, 4)
 
Output: true
Explanation: One possible way is : left -> down -> left -> down -> right -> down -> right.

Example 2

Input 1: a maze represented by a 2D array
 
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
 
Input 2: start coordinate (rowStart, colStart) = (0, 4)
Input 3: destination coordinate (rowDest, colDest) = (3, 2)
 
Output: false
Explanation: There is no way for the ball to stop at the destination.

Note:

1. There is only one ball and one destination in the maze.
2. Both the ball and the destination exist on an empty space, and they will not be at the same position initially.
3. The given maze does not contain border (like the red rectangle in the example pictures), but you could assume the border of the maze are all walls.
4. The maze contains at least 2 empty spaces, and both the width and height of the maze won't exceed 100.

这个题目思路就是用DFS/BFS, 需要注意的是每次append某个方向时要加入一个while loop使得一直碰到边缘或者墙壁, 也就是'1' 为止, 然后以此作为条件去判断是否append进入

stack或者queue里面.

1. Constraints

1) 基本都是valid

2. Ideas

DFS/BFS      T; O(m*n)    S: O(m*n)

3. Code

class Solution:
    def maze(self, maze, start, destination):
        lr, lc, visited = len(maze), len(maze[0]), set([tuple(start)])
        queue = collections.deque([tuple(start)])
       # stack = [tuple(start)]    # DFS, stack.pop() 去代替queue.popleft()即可
        dirs = [(0,-1), (0,1), (-1,0), (1,0)]
        while queue:
            pr, pc = queue.popleft()
            if [pr,pc] == destination:
                return True
            for c1, c2 in dirs:
                nr, nc = pr + c1, pc + c2
                while 0 <= nr <lr and 0 <= nc < lc and maze[nr][nc] != 1:
                    nr += c1
                    nc += c2
                nr -= c1
                nc -= c2
                if maze[nr][nc] == 0 and (nr, nc) not in visited:
                    queue.append((nr, nc))
                    visited.add((nr, nc))
         return False