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.
有一個球在一個二維數組表示的空間里，1代表是牆，0代表是空，邊界可以看作是牆。球可以向上下左右四個方向滾動，每一次滾動到牆才停止然后進行下一次滾動。求球是否可以滾動到終點。
解法1：DFS，可能TLE。首先判斷是否已經到達了終點，如果是則直接返回true；否則就看該位置是否已經被訪問過了，如果是則返回，否則就記當前位置已經被訪問。然后用DFS嘗試四個不同方向，有任何一個方向可以到達終點就直接返回true。
解法2: BFS。建立一個queue，先將start位置放入隊列。每次從隊列頭部拿出一個位置，如果此位置是終點則直接返回true，否則就看該位置是否被訪問過，如果沒有被訪問，將其四個方向上移動可以停靠的點加入隊列，此位置記為已訪問。如果隊列空了還沒有發現可以到達的路徑，說明無法到達了。
Java:
public class Solution {
public boolean hasPath(int[][] maze, int[] start, int[] destination) {
if(maze.length == 0 || maze[0].length == 0) return false;
if(start[0] == destination[0] && start[1] == destination[1]) return true;
m = maze.length; n = maze[0].length;
boolean[][] visited = new boolean[m][n];
return dfs(maze, start, destination, visited);
}
int m, n;
int[][] dirs = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
private boolean dfs(int[][] maze, int[] cur, int[] dest, boolean[][] visited) {
// already visited
if(visited[cur[0]][cur[1]]) return false;
// reach destination
if(Arrays.equals(cur, dest)) return true;
visited[cur[0]][cur[1]] = true;
for(int[] dir : dirs) {
int nx = cur[0], ny = cur[1];
while(notWall(nx + dir[0], ny + dir[1]) && maze[nx+dir[0]][ny+dir[1]] != 1) {
nx += dir[0]; ny += dir[1];
}
if(dfs(maze, new int[] {nx, ny}, dest, visited)) return true;
}
return false;
}
private boolean notWall(int x, int y) {
return x >= 0 && x < m && y >= 0 && y < n;
}
}
Python:
class Solution(object): def hasPath(self, maze, start, destination): """ :type maze: List[List[int]] :type start: List[int] :type destination: List[int] :rtype: bool """ start = tuple(start) destination = tuple(destination) if start == destination: return True directions = [(0, 1), (0, -1), (1, 0), (-1, 0)] width = len(maze) height = len(maze[0]) if 1 <= destination[0] < width-1 and 1 <= destination[1] < height-1: if not any([maze[destination[0]+direction[0]][destination[1]+direction[1]] for direction in directions]): return False positions = [start] visited = set() while positions: position = positions.pop(0) for direction in directions: if (position + direction) in visited: continue nextPosition = position while (nextPosition+direction) not in visited: visited.add(nextPosition+direction) prevPosition = nextPosition nextPosition = (nextPosition[0]+direction[0], nextPosition[1]+direction[1]) if not 0 <= nextPosition[0] < width or not 0 <= nextPosition[1] < height or maze[nextPosition[0]][nextPosition[1]] == 1: if prevPosition == destination: return True positions.append(prevPosition) break return False
C++: DFS
class Solution {
public:
vector<vector<int>> dirs{{0,-1},{-1,0},{0,1},{1,0}};
bool hasPath(vector<vector<int>>& maze, vector<int>& start, vector<int>& destination) {
if (maze.empty() || maze[0].empty()) return true;
int m = maze.size(), n = maze[0].size();
vector<vector<int>> dp(m, vector<int>(n, -1));
return helper(maze, dp, start[0], start[1], destination[0], destination[1]);
}
bool helper(vector<vector<int>>& maze, vector<vector<int>>& dp, int i, int j, int di, int dj) {
if (i == di && j == dj) return true;
if (dp[i][j] != -1) return dp[i][j];
bool res = false;
int m = maze.size(), n = maze[0].size();
maze[i][j] = -1;
for (auto dir : dirs) {
int x = i, y = j;
while (x >= 0 && x < m && y >= 0 && y < n && maze[x][y] != 1) {
x += dir[0]; y += dir[1];
}
x -= dir[0]; y -= dir[1];
if (maze[x][y] != -1) {
res |= helper(maze, dp, x, y, di, dj);
}
}
dp[i][j] = res;
return res;
}
};
C++: BFS
class Solution {
public:
bool hasPath(vector<vector<int>>& maze, vector<int>& start, vector<int>& destination) {
if (maze.empty() || maze[0].empty()) return true;
int m = maze.size(), n = maze[0].size();
vector<vector<bool>> visited(m, vector<bool>(n, false));
vector<vector<int>> dirs{{0,-1},{-1,0},{0,1},{1,0}};
queue<pair<int, int>> q;
q.push({start[0], start[1]});
visited[start[0]][start[1]] = true;
while (!q.empty()) {
auto t = q.front(); q.pop();
if (t.first == destination[0] && t.second == destination[1]) return true;
for (auto dir : dirs) {
int x = t.first, y = t.second;
while (x >= 0 && x < m && y >= 0 && y < n && maze[x][y] == 0) {
x += dir[0]; y += dir[1];
}
x -= dir[0]; y -= dir[1];
if (!visited[x][y]) {
visited[x][y] = true;
q.push({x, y});
}
}
}
return false;
}
};
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