leetcode 【 Unique Paths II 】 python 实现

题目：

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.

代码：oj测试通过 Runtime: 48 ms

class Solution:
# @param obstacleGrid, a list of lists of integers
# @return an integer
def uniquePathsWithObstacles(self, obstacleGrid):
# ROW & COL
ROW = len(obstacleGrid)
COL = len(obstacleGrid[0])
# one row case
if ROW==1:
for i in range(COL):
if obstacleGrid[0][i]==1:
return 0
return 1
# one column case
if COL==1:
for i in range(ROW):
if obstacleGrid[i][0]==1:
return 0
return 1
# visit normal case
dp = [[0 for i in range(COL)] for i in range(ROW)]
for i in range(COL):
if obstacleGrid[0][i]!=1:
dp[0][i]=1
else:
break
for i in range(ROW):
if obstacleGrid[i][0]!=1:
dp[i][0]=1
else:
break
# iterator the other nodes
for row in range(1,ROW):
for col in range(1,COL):
if obstacleGrid[row][col]==1:
dp[row][col]=0
else:
dp[row][col]=dp[row-1][col]+dp[row][col-1]

return dp[ROW-1][COL-1]

思路：

思路模仿Unique Path这道题：

/article/5261261.html

只不过多了某些障碍点；针对障碍点，加一个判断条件即可：如果遇上障碍点，那么到途径这个障碍点到达终点的可能路径数为0。然后继续迭代到尾部即可。

个人感觉40行的python脚本不够简洁，总是把special case等单独拎出来。后面再练习代码的时候，考虑如何让代码更简洁。