LeetCode 63 不同路径II

题目描述：

思路：

基本思路还是和62题不同路径相似，除0行和0列之外，其他位置的值是上侧和左侧值之和。

不同的是要加入查障碍，先将矩阵中的障碍1转化为-1，若0行或0列有-1，此位置之后的所有值均为-1。更新其他位置时，如果此位置为-1则不做更新，如果上左有一个-1，则只加另一侧值，上

class Solution:
    def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
        if len(obstacleGrid) == 0:
            return None
        if obstacleGrid[0][0] == 1 or obstacleGrid[len(obstacleGrid)-1][len(obstacleGrid[0])-1] == 1:
            return 0
        for i in range(len(obstacleGrid)):
            for j in range(len(obstacleGrid[0])):
                if obstacleGrid[i][j] == 1:
                    obstacleGrid[i][j] = -1
        obstacleGrid[0][0] = 1
        #print(tem)
        for i in range(len(obstacleGrid)):
            for j in range(len(obstacleGrid[0])):
                if obstacleGrid[i][j] == -1:
                    continue
                if i == 0 and j == 0:
                    continue
                elif i == 0 and j != 0:
                    obstacleGrid[i][j] = obstacleGrid[i][j-1]
                    #print('1',tem[i-1][j])
                elif j == 0 and i != 0:
                    obstacleGrid[i][j] = obstacleGrid[i-1][j]
                else:
                    if obstacleGrid[i-1][j] == -1 and obstacleGrid[i][j-1] != -1:
                        obstacleGrid[i][j] += obstacleGrid[i][j-1]
                    elif obstacleGrid[i-1][j] != -1 and obstacleGrid[i][j-1] == -1:
                        obstacleGrid[i][j] += obstacleGrid[i-1][j]
                    elif obstacleGrid[i-1][j] == -1 and obstacleGrid[i][j-1] == -1:
                        obstacleGrid[i][j] == -1
                    else:
                        obstacleGrid[i][j] = obstacleGrid[i-1][j] + obstacleGrid[i][j-1]
                #print('i',i,'j',j,'  ',tem)
        #print(tem)
        return obstacleGrid[len(obstacleGrid)-1][len(obstacleGrid[0])-1]

左都是-1，此位置也为-1。