####问题描述
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:
There is only one ball and one destination in the maze.
Both the ball and the destination exist on an empty space, and they will not be at the same position initially.
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.
The maze contains at least 2 empty spaces, and both the width and height of the maze won’t exceed 100.
题目链接：
####思路分析
给一二维数组表示一个迷宫，球只有碰到墙壁之后才会停下，问球能否到达终点。
For this question, it is a variation of ordinary dfs problem. Because in this question, the ball could move forward through multiple cells not one.
####代码
class Solution {
private int[][] dirs = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
public boolean hasPath(int[][] maze, int[] start, int[] destination) {
if (maze.length == 0 || maze[0].length == 0){
return false;
}
return dfs(maze, start, destination, new boolean[maze.length][maze[0].length]);
}
private boolean dfs(int[][] maze, int[] current, int[] destination, boolean[][] visited){
int x = current[0];
int y = current[1];
if (x == destination[0] && y == destination[1]){
return true;
}
if (x < 0 || y < 0 || x >= maze.length || y >= maze[0].length || visited[x][y]){
return false;
}
visited[x][y] = true;
for (int[] dir : dirs){
int xx = x, yy = y;
while (xx >= 0 && xx < maze.length && yy >= 0 && yy < maze[0].length && maze[xx][yy] == 0 ){
xx += dir[0];
yy += dir[1];
}
//retrive one step for the while loop is moving forward one cell.
if (dfs(maze, new int[]{xx - dir[0], yy - dir[1]}, destination, visited)){
return true;
}
}
return false;
}
}
时间复杂度：
O
(
n
)
O(n)
O
(
n
)
空间复杂度：
O
(
n
)
O(n)
O
(
n
)
反思