这道题使用dfs会超时，看评论区也有人遇到同样的问题，比赛时调试了1个多小时尝试改进，没有意识到应该换用非递归的bfs可以解决，消耗了大量的时间。
超时的方案如下，使用python实现：（经过尝试，能通过的测试用例中，使用5个方向就可以了）
1 import sys 2 class Solution: 3 def __init__(self): 4 self.visited = [] 5 self.distance = sys.maxsize 6 #self.direction = [[-1,-1],[-1,0],[-1,1],[0,-1],[0,1],[1,-1],[1,0],[1,1]] 7 self.direction = [[1,1],[0,1],[1,0],[-1,1],[1,-1]] 8 #self.direction = [[1,1],[0,1],[1,0],[1,-1]] 9 #self.direction = [[1,1],[0,1],[1,0]] 10 11 def dfs(self,grid,N,i,j,distance): 12 if i == N-1 and j == N-1: 13 self.distance = min(self.distance,distance+1) 14 return 15 16 if distance >= self.distance: 17 return 18 19 if self.visited[i][j] == 0: 20 self.visited[i][j] = 1 21 distance += 1 22 for direct in self.direction: 23 x = i + direct[0] 24 y = j + direct[1] 25 if x < 0 or x >= N or y < 0 or y >= N: 26 continue 27 if grid[x][y] == 1 or self.visited[x][y] == 1: 28 continue 29 self.dfs(grid,N,x,y,distance) 30 distance -= 1 31 self.visited[i][j] = 0 32 33 34 def shortestPathBinaryMatrix(self, grid: 'List[List[int]]') -> int: 35 N = len(grid) 36 if grid[0][0] == 1 or grid[N-1][N-1] == 1: 37 return -1 38 self.visited = [[0 for _ in range(N)]for _ in range(N)] 39 40 self.dfs(grid,N,0,0,0) 41 return self.distance if self.distance < sys.maxsize else -1
下面是AC的解决方案，使用bfs思想，代码如下：
1 import sys 2 class Solution: 3 def __init__(self): 4 self.visited = [] 5 self.distance = sys.maxsize 6 self.direction = [[1,1],[0,1],[1,0],[-1,1],[1,-1]] 7 8 def bfs(self,grid,queue,N): 9 while len(queue) != 0: 10 n = len(queue) 11 while n > 0: 12 cur = queue.pop(0) 13 for direct in self.direction: 14 x = cur[0] + direct[0] 15 y = cur[1] + direct[1] 16 distance = self.visited[cur[0]][cur[1]] 17 if x == N-1 and y == N-1: 18 self.distance = min(self.distance,distance + 1) 19 continue 20 if x < 0 or x >= N or y < 0 or y >= N: 21 continue 22 if grid[x][y] == 1: 23 continue 24 25 if self.visited[x][y] == 0: 26 self.visited[x][y] = distance + 1 27 elif self.visited[x][y] > distance + 1: 28 self.visited[x][y] = distance + 1 29 else: 30 continue 31 queue.append([x,y]) 32 n = len(queue) 33 34 def shortestPathBinaryMatrix(self, grid: 'List[List[int]]') -> int: 35 N = len(grid) 36 if grid[0][0] == 1 or grid[N-1][N-1] == 1: 37 return -1 38 if N == 1: 39 return 1 40 41 self.visited = [[0 for _ in range(N)]for _ in range(N)] 42 43 queue = [[0,0]] 44 self.visited[0][0] = 1 45 self.bfs(grid,queue,N) 46 47 return self.distance if self.distance < sys.maxsize else -1