class Solution(object): def uniquePathsWithObstacles(self, obstacleGrid): """ :type obstacleGrid: List[List[int]] :rtype: int """ m = len(obstacleGrid) n = len(obstacleGrid) 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
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
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