63. 不同路径 II

63. 不同路径 II
-> 就前一题而言，在其基础上添加判断条件，当遇到障碍时，置为0并在边界条件下也需要做判断，话不多说
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class Solution:
    def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
        row = len(obstacleGrid)
        col = len(obstacleGrid[0])
        dp = [[0 for _ in range(col)] for _ in range(row)]
        for j in range(col):#应该从起点0开始，我开始写的1，col
            if obstacleGrid[0][j]!=1:
                dp[0][j] = 1 #边界复制,此时需做判断是否为障碍
            else:
                break
        for i in range(row):
            if obstacleGrid[i][0]!=1:
                dp[i][0]=1
            else:
                break
        for i in range(1,row):
            for j in range(1,col):
                if obstacleGrid[i][j]==1:
                    dp[i][j]=0 #若遇到障碍，将其置为0
                else:
                    dp[i][j] = dp[i-1][j]+dp[i][j-1]
        return  dp[-1][-1]