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leetcode之Unique Paths II

2016-01-08 01:25 281 查看
这题相比于第一题就是增加了阻碍。不过我做I的时候的思路是以右下角为0,0,导致多做了好几遍无用功。反而速度不快了。可以改进下的其实。代码如下:
class Solution(object):
def uniquePathsWithObstacles(self, obstacleGrid):
"""
:type obstacleGrid: List[List[int]]
:rtype: int
"""
if obstacleGrid[len(obstacleGrid) - 1][len(obstacleGrid[0]) - 1] == 1:
return 0
for i in range(len(obstacleGrid)):
for j in range(len(obstacleGrid[0])):
if obstacleGrid[i][j] == 0:
obstacleGrid[i][j] = 1
else:
obstacleGrid[i][j] = 0
#去除边界的阻碍的情况
for i in range(len(obstacleGrid))[::-1]:
if obstacleGrid[i][-1] == 0:
for j in range(i):
obstacleGrid[j][-1] = 0
break

for i in range(len(obstacleGrid[0]))[::-1]:
if obstacleGrid[-1][i] == 0:
for j in range(i):
print j, i
obstacleGrid[-1][j] = 0
break
#去除中间有阻碍的情况
for i in range(len(obstacleGrid) - 1)[::-1]:
for j in range(len(obstacleGrid[0]) - 1)[::-1]:
if obstacleGrid[i][j] != 0:
if obstacleGrid[i + 1][j] != 0 and obstacleGrid[i][j + 1] != 0:
obstacleGrid[i][j] = obstacleGrid[i][j + 1] + obstacleGrid[i + 1][j]
elif obstacleGrid[i + 1][j] == 0 and obstacleGrid[i][j + 1] == 0:
obstacleGrid[i][j] = 0
elif obstacleGrid[i + 1][j] == 0:
obstacleGrid[i][j] = obstacleGrid[i][j + 1]
else:
obstacleGrid[i][j] = obstacleGrid[i + 1][j]
return obstacleGrid[0][0]
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