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
.
Note: m and n will be at most 100.
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与之前的unique path 很像,就是需要对每一个砖进行判断,而且初始化也需要重新写
class Solution(object):
def uniquePathsWithObstacles(self, obstacleGrid):
x = obstacleGrid
m = len(x)
n = len(x[0])
r1 = [0] * n
r2 = [0] * n
for i in xrange(len(r1)):
if x[0][i] == 1:
break
r1[i] = 1
r2 = r1
for i in range(1,m):
r2[0] = r1[0]*(1 - x[i][0])
for j in range(1,n):
r2[j] = r1[j]*(1 - x[i][j]) + r2[j -1]*(1 - x[i][j])
r1 = r2
#print r2
return r2[n-1]