L 490. The Maze ----M 我当年的Google on site题

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 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:

• 1 <= maze.length, maze[i].length <= 100
• maze[i][j] is 0 or 1.
• start.length == 2
• destination.length == 2
• 0 <= startrow, destinationrow <= maze.length
• 0 <= startcol, destinationcol <= maze[i].length
• Both the ball and the destination exist on an empty space, and they will not be at the same position initially.
• The maze contains at least 2 empty spaces.

 A:

DFS

就是每次不直接找到下一个节点。 而是找到所有的靠墙的节点

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class Solution { public: bool hasPath(vector<vector<int>>& maze, vector<int>& start, vector<int>& destination) { int m = maze.size(); int n = maze[0].size(); if(start == destination) return true; if(maze[start[0]][start[1]] !=0){ return false; } maze[start[0]][start[1]] = -1; vector<vector<int>> Diff{{1,0},{-1,0}, {0,1}, {0,-1}}; for(auto d : Diff){ auto pre = start; while(true){ auto next = vector<int>{pre[0]+d[0], pre[1] + d[1]}; if(next[0]>= 0 && next[0] <m && next[1] >=0 & next[1] < n && maze[next[0]][next[1]] != 1){ pre = next; }else{ break; } } if(maze[pre[0]][pre[1]] == 0){ auto res = hasPath(maze, pre, destination); if(res) return true; } } return false; } };

犯的错误：

一开始没有看懂题意，还 以为是所有经过的都可以呢。

***************  solution 2 ****既然只能靠墙的才能算，那我们只grow*********