[leetcode] Unique Path II

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.

思路： unique path 1 很trivial了，直接动态规划，2的话要考虑特殊情况，比如第一行和第一列初始化的时候，一旦遇到1，则后面的都无法到达，则后面的点都是0。再计算其他点的时候，考虑上面或者左面存在0点的时候（0表示0种方法到达该点）。

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