class Solution(object):
def uniquePathsWithObstacles(self, obstacleGrid):
"""
:type obstacleGrid: List[List[int]]
:rtype: int
"""
m = len(obstacleGrid)
n = len(obstacleGrid[0])
mat = [[0 for _ in range(n+1)] for _ in range(m+1)]
# mat[i][j] represents the number of ways for (i+,j+1) to (m,n),
# so mat[m-1][n-1] = 1
for i in range(m-1,-1,-1):
for j in range(n-1,-1,-1):
if obstacleGrid[i][j] == 1:
mat[i][j] = 0
elif i == m-1 and j == n-1:
mat[i][j] = 1
else:
mat[i][j] = mat[i+1][j] + mat[i][j+1]
return mat[0][0]
https://leetcode.com/problems/unique-paths-ii/
Follow up for "Unique Paths":
Now consider if some obstacles are added to the grids. How many unique paths would there be?
An obstacle and empty space is marked as 1
and 0
respectively in the grid.
For example, There is one obstacle in the middle of a 3x3 grid as illustrated below.
[
[0,0,0],
[0,1,0],
[0,0,0]
]
The total number of unique paths is 2
.