# 2D DP, 本题也可以写成1D DP（滚动数组）的形式减少空间复杂度
class Solution:
def uniquePaths(self, m: int, n: int) -> int:
dp = [[1] * n for _ in range(m)]
for i in range(1, m):
for j in range(1, n):
dp[i][j] = dp[i-1][j] + dp[i][j-1]
return dp[-1][-1]
class Solution:
def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
n_row = len(obstacleGrid)
n_col = len(obstacleGrid[0])
if obstacleGrid[n_row - 1][n_col - 1] == 1 or obstacleGrid[0][0] == 1:
return 0
res = [[0 for _ in range(n_col)] for _ in range(n_row)]
for i in range(n_row):
if obstacleGrid[i][0] == 1:
break
res[i][0] = 1
for i in range(n_col):
if obstacleGrid[0][i] == 1:
break
res[0][i] = 1
for i in range(1, n_row):
for j in range(1, n_col):
if obstacleGrid[i][j] == 1:
continue
else:
res[i][j] = res[i-1][j] + res[i][j-1]
return res[-1][-1]