Note: m and n will be at most 100.

这题的做法和前面的unique path是一样的，都是用的动态规划。不同之处在于需要判断有没有点为1.

直接上代码：

class Solution:
    # @param obstacleGrid, a list of lists of integers
    # @return an integer
    def uniquePathsWithObstacles(self, obstacleGrid):
        rown=len(obstacleGrid)
        coln=len(obstacleGrid[0])
        dp=[[0 for i in range(coln)] for j in range(rown)]
        if obstacleGrid[0][0]!=1:
            dp[0][0]=1
        for row in range(1,rown):
            if obstacleGrid[row][0]!=1 and dp[row-1][0]==1:
                dp[row][0]=1
            else:
                break
        for col in range(1,coln):
            if obstacleGrid[0][col]!=1 and dp[0][col-1]==1:
                dp[0][col]=1
            else:
                break
        for row in range(1,rown):
            for col in range(1,coln):
                if obstacleGrid[row][col]==1:
                    dp[row][col]=0
                else:
                    dp[row][col]=dp[row-1][col]+dp[row][col-1]
        return dp[rown-1][coln-1]

63. Unique Path II Leetcode Python