# my solution with explanation

• ``````    public boolean hasPath(int[][] maze, int[] start, int[] destination) {
int m = maze.length;
int n = maze[0].length;
boolean[][] visited = new boolean[m][n];
return helper(maze, visited, start[0], start[1], destination[0], destination[1]);
}

public boolean helper(int[][] maze, boolean[][] visited, int curX, int curY, int desX, int desY){
if(visited[curX][curY]) return false;
visited[curX][curY] = true;
if(curX == desX && curY == desY) return true;
int[][] directions = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
for(int[] dir: directions){
int tmpX = curX, tmpY = curY;
while(tmpX + dir[0] >= 0 && tmpX + dir[0] < maze.length && tmpY + dir[1] >= 0 && tmpY + dir[1] < maze[0].length && maze[tmpX +  dir[0]][tmpY + dir[1]] == 0){
tmpX += dir[0];
tmpY += dir[1];
}
if(tmpX != curX || tmpY != curY){
if(helper(maze, visited, tmpX, tmpY, desX, desY)) return true;
}
}
// visited[curX][curY] = true;
return false;
}
}
``````

this question is not that hard to think about but has some corner cases to consider and need to think deeply about the function stack,
if I put visited[][] = true at the slashed place, it will cause stack overflow.

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