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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
1 class Solution: 2 # @param obstacleGrid, a list of lists of integers 3 # @return an integer 4 def uniquePathsWithObstacles(self, obstacleGrid): 5 # ROW & COL 6 ROW = len(obstacleGrid) 7 COL = len(obstacleGrid[0]) 8 # one row case 9 if ROW==1:10 for i in range(COL):11 if obstacleGrid[0][i]==1:12 return 013 return 114 # one column case15 if COL==1:16 for i in range(ROW):17 if obstacleGrid[i][0]==1:18 return 019 return 120 # visit normal case21 dp = [[0 for i in range(COL)] for i in range(ROW)]22 for i in range(COL):23 if obstacleGrid[0][i]!=1:24 dp[0][i]=125 else:26 break27 for i in range(ROW):28 if obstacleGrid[i][0]!=1:29 dp[i][0]=130 else:31 break32 # iterator the other nodes33 for row in range(1,ROW):34 for col in range(1,COL):35 if obstacleGrid[row][col]==1:36 dp[row][col]=037 else:38 dp[row][col]=dp[row-1][col]+dp[row][col-1]39 40 return dp[ROW-1][COL-1]
思路：
思路模仿Unique Path这道题：
http://www.cnblogs.com/xbf9xbf/p/4250359.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。
leetcode 【 Unique Paths II 】 python 实现
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