LeetCode--Unique Paths II

题目：

接着上一道题目Unique Paths，考虑到障碍问题。在matrix（m*n）矩阵中有障碍的地方设为1，否则为0。计算到达右下角的不同路径有多少种。

解题思路：

考虑使用动态规划的方式解题，同上个题目一样，要到达（i，j），则要先到达（i-1，j）或者（i，j-1）。另外若（i，j）位置为障碍物的话，到达（i，j）的路径个数为0.

代码（python）：

class Solution(object):
def uniquePathsWithObstacles(self, obstacleGrid):
"""
:type obstacleGrid: List[List[int]]
:rtype: int
"""
for i in range(len(obstacleGrid)):
for j in range(len(obstacleGrid[0])):
if obstacleGrid[i][j]==1:
obstacleGrid[i][j]=0
continue
elif i==0 and j==0:
obstacleGrid[i][j]=1
continue
elif i==0 and j!=0:
obstacleGrid[i][j]=obstacleGrid[i][j-1]
continue
elif i!=0 and j==0:
obstacleGrid[i][j]=obstacleGrid[i-1][j]
continue
else:
obstacleGrid[i][j]=obstacleGrid[i-1][j]+obstacleGrid[i][j-1]
return obstacleGrid[len(obstacleGrid)-1][len(obstacleGrid[0])-1]