解1 递归

dfs深度优先
超出时间限制

class Solution:
def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:

if obstacleGrid == []:
return 0

if obstacleGrid[0][0] == 1:
return 0

m = len(obstacleGrid)
n = len(obstacleGrid[0])
if obstacleGrid[m - 1][n - 1] == 1:
return 0

result = list()
self.dfs(obstacleGrid, 0, 0, result)
return sum(result)

def dfs(self, obstacleGrid, i, j, result):

if i == len(obstacleGrid) - 1 and j == len(obstacleGrid[0]) - 1:
result.append(1)

elif i == len(obstacleGrid) - 1:
if obstacleGrid[i][j + 1] == 1:
return
else:
self.dfs(obstacleGrid, i, j + 1, result)

elif j == len(obstacleGrid[0]) - 1:
if obstacleGrid[i + 1][j] == 1:
return
else:
self.dfs(obstacleGrid, i + 1, j, result)

else:
if obstacleGrid[i][j + 1] == 0:
self.dfs(obstacleGrid, i, j + 1, result)

if obstacleGrid[i + 1][j] == 0:
self.dfs(obstacleGrid, i + 1, j, result)

解2 动态规划

参考62

class Solution:
def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:

if obstacleGrid == []:
return 0

# 机器人所在位置为障碍物 无法开始
if obstacleGrid[0][0] == 1:
return 0

# 终点为障碍物 无法到达
m = len(obstacleGrid)
n = len(obstacleGrid[0])
if obstacleGrid[m - 1][n - 1] == 1:
return 0

# 动态规划
path = [[1 for i in range(n)] for j in range(m)]

# 若第1列 有一个方格为障碍物 那么相当于它下面的方格都为障碍物
for i in range(1, m):
if obstacleGrid[i-1][0]:
obstacleGrid[i][0] = 1

# 同理 第一行有一个为障碍物 右边的方格都为障碍物
for j in range(1, n):
if obstacleGrid[0][j - 1]:
obstacleGrid[0][j] = 1

for i in range(1, m):
for j in range(1, n):

if (obstacleGrid[i - 1][j] and obstacleGrid[i][j - 1]) or obstacleGrid[i][j]:
# 上 左 都无法到达 或 本身就是障碍
path[i][j] = 0

elif obstacleGrid[i - 1][j]:
# 上 为障碍物
path[i][j] = path[i][j - 1]

elif obstacleGrid[i][j - 1]:
# 左 为障碍物
path[i][j] = path[i - 1][j]

else:
# 上 左 都可到达
path[i][j] = path[i][j - 1] + path[i - 1][j]

return path[m - 1][n - 1]