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
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
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.
解法1：DFS + Memorization
这题和一般的地图题或者是有一题岛屿的题目类似，都可以用DFS来解决。这里的难点或者是区别在，球会一直往一个方向滚直到撞到了墙。所以在设计dfs程序的时候要考虑这个问题，可以用一个while循环解决。另外，对于已经访问过的点，不需要额外的申请一个空间来存储，只需要把已经访问过的node对应的maze的值设为-1就可。
对于dfs的优化呢，可以用一个dp矩阵来记录每一个扫描过的初始节点是否有解来剪掉一些枝
Java
public class Solution {
public boolean hasPath(int[][] maze, int[] start, int[] destination) {
if (maze.length == 0 || maze[0].length == 0) {
return true;
}
int[][] dp = new int[maze.length][maze[0].length];
for (int[] row: dp) {
Arrays.fill(row, -1);
}
return dfs(maze, start[0], start[1], destination, dp);
}
private boolean dfs(int[][] maze, int i, int j, int[] destination, int[][] dp) {
if (i == destination[0] && j == destination[1]) {
return true;
}
if (dp[i][j]!= -1) {
return dp[i][j] == 1 ? true: false;
}
boolean res = false;
int[][] directions = {{0,-1},{-1,0},{0,1},{1,0}};
maze[i][j] = -1;
for (int[] dir: directions) {
int x = i, y = j;
int xinc = dir[0];
int yinc = dir[1];
while (x >= 0 && x < maze.length && y >= 0 && y < maze[0].length && maze[x][y] != 1) {
x += xinc;
y += yinc;
}
x -= xinc;
y -= yinc;
if (maze[x][y] != -1) {
res = res || dfs(maze, x,y, destination, dp);
}
}
dp[i][j] = res? 1: 0;
maze[i][j] = 0;
return res;
}
}